NDA Permutations and Combinations previous year questions with solutions

Answer: (a) 362880

Method. solve (2x)!/⁠x! = 120 by trial

z/⁠y = (2x)!/⁠x! = (x + 1)(x + 2)…(2x).
x = 3: 4 × 5 × 6 = 120 (holds) (x = 2 gives 12, x = 4 gives 1680, and the product increases with x).
(3x)! = 9! = 362880.

Answer: (c) 5

Method. trailing zeros of n! = [n/⁠5] + [n/⁠25] + …

For n < 25 the number of trailing zeros of n! is [n/⁠5].
[n/⁠5] = 2 exactly when 10 ≤ n ≤ 14; for n ≥ 15 there are at least 3 zeros, and for n ≤ 9 at most 1.
So n = 10, 11, 12, 13, 14: 5 values.

Answer: (b) 7

Method. Legendre's formula for the power of 2 in 26!

Power of 2 in 26! = [26/⁠2] + [26/⁠4] + [26/⁠8] + [26/⁠16] = 13 + 6 + 3 + 1 = 23.
8^k = 2^3k divides 26! only if 3k ≤ 23, i.e. k ≤ 7.
So the maximum value of k is 7 (with n = 26!/⁠8⁷, a positive integer).

Answer: (c) 7

Method. k! is divisible by 24 for k ≥ 4

For k ≥ 4, k! contains 4! = 24, so 5!, 7!, …, 199! each leave remainder 0 on division by 24.
Remainder = remainder of 1! + 3! = 1 + 6 = 7.

Answer: (b) 2 only

Method. Wilson's theorem holds only for prime moduli

Statement 1: 26 = 2 × 13 and both 2 and 13 occur in 25!, so 26 divides 25!; then 25! + 1 leaves remainder 1, so it is false.
Statement 2: 6! + 1 = 721 = 7 × 103, so it is true (Wilson: (p − 1)! + 1 is divisible by prime p).
Answer: 2 only.

Answer: (c) 2

Method. k! is divisible by 8 for k ≥ 4

For k ≥ 4, k! contains 4! = 24 = 8 × 3, so it leaves remainder 0 on division by 8.
Remainder of S = remainder of 0! + 1! + 2! + 3! = 1 + 1 + 2 + 6 = 10.
10 = 8 + 2, so the remainder is 2.

Answer: (d) 34

Method. k! is divisible by 60 for k ≥ 5

For k ≥ 5, k! contains 5! = 120 = 2 × 60, so it leaves remainder 0 on division by 60.
Remainder of S = remainder of 0! + 1! + 2! + 3! + 4! = 1 + 1 + 2 + 6 + 24 = 34.
34 < 60, so the remainder is 34.

Answer: (c) 7

Method. factor out 30!, then count factors of 5 by Legendre's formula

30! + 35! = 30!(1 + 31 × 32 × 33 × 34 × 35).
The bracket is 1 + (a multiple of 35), so it leaves remainder 1 on division by 5 and has no factor 5.
Power of 5 in 30! = ⌊305⌋ + ⌊3025⌋ = 6 + 1 = 7.
Maximum n = 7.

Answer: (b) 10!

Method. telescoping: k · k! = (k + 1)! − k!

The sum is 9Σk=2 k · k! + 2.
k · k! = (k + 1 − 1) · k! = (k + 1)! − k!.
9Σk=2 [(k + 1)! − k!] = 10! − 2! = 10! − 2.
Adding 2 gives 10!.

Answer: (d) Neither 1 nor 2

Method. test the smallest admissible values of n

n!3! = 4 × 5 × … × n for n > 3.
Statement 1: n = 4 gives 4!3! = 4, not divisible by 6, so it fails.
Statement 2: n = 4 gives 4 + 3 = 7 (divisible by 7), but n = 5 gives 5!3! + 3 = 20 + 3 = 23, not divisible by 7, so it fails.
Neither statement holds for every n > 3.

Answer: (d) No such value of n exists

Method. trailing zeros of n! = ⌊n/⁠5⌋ + ⌊n/⁠25⌋ + …

The number of zeros at the end of n! is Z(n) = ⌊n/⁠5⌋ + ⌊n/⁠25⌋ + ⌊n/⁠125⌋ + ….
Z(74) = 14 + 2 = 16 and Z(75) = 15 + 3 = 18, and Z(n) never decreases as n grows.
So Z(n) jumps from 16 straight to 18 at n = 75 (75 = 3 × 25 adds two 5s at once): no n! ends in exactly 17 zeros.
Check the options: Z(80) = 19, Z(85) = 20, Z(95) = 22.
Hence no such value of n exists.

'Has 17 zeros' is read, as the options require, as 'ends in 17 zeros' (the factorial trailing-zero problem); counting every zero digit of n! would give n = 53, 56 or 58, none of which is an option.

Answer: (c) 232

Method. complement: all numbers minus those with distinct digits

With repetition allowed: 4⁴ = 256 numbers.
With all digits distinct: 4! = 24.
At least one digit repeated: 256 − 24 = 232.

Answer: (b) 18

Method. only 4-digit numbers can exceed 1000

With 4 distinct digits the numbers have at most 4 digits, and a number with 3 or fewer digits is below 1000.
4-digit numbers: first digit from 1, 2, 3 (3 ways), remaining three digits in 3! = 6 ways: 3 × 6 = 18.
The smallest is 1023 > 1000, so all 18 count.

Answer: (b) 6

Method. divisibility by 4 depends on the last two digits

Endings from 1, 2, 3, 4 with distinct digits divisible by 4: 12, 24, 32.
For each ending the other two digits fill the first two places in 2! = 2 ways.
Number = 3 × 2 = 6.

Answer: (a) 625

Method. multiplication principle

Odd digits: 1, 3, 5, 7, 9 (none is 0, so the first place has no restriction).
Each of the four places has 5 choices: 5⁴ = 625.

Answer: (d) 64440

Method. place-value sum over all arrangements, minus those with leading 0

All 4! = 24 arrangements of 0, 1, 4, 5: each digit occurs 6 times in each place, so their sum = 6 × (0 + 1 + 4 + 5) × 1111 = 6 × 10 × 1111 = 66660.
Arrangements with 0 first are really 3-digit numbers from 1, 4, 5: each digit occurs 2 times in each place, sum = 2 × 10 × 111 = 2220.
Sum of the four-digit numbers = 66660 − 2220 = 64440.

Answer: (b) 500

Method. multiplication principle with a non-zero leading digit

Even digits: 0, 2, 4, 6, 8.
Thousands digit cannot be 0: 4 choices (2, 4, 6, 8); each of the other three digits: 5 choices.
Number = 4 × 5 × 5 × 5 = 500.

Answer: (d) 6

Method. divisibility by 4 depends on the last two digits

A number is divisible by 4 when the number formed by its last two digits is divisible by 4.
Two-digit endings from 1, 2, 3, 5 with distinct digits that are divisible by 4: 12, 32, 52 (an ending in 1, 3 or 5 is odd; 12, 32, 52 are the only ones ending in 2).
For each ending the remaining two digits fill the first two places in 2! = 2 ways.
Number = 3 × 2 = 6.

Answer: (d) 312

Method. divisible by 6 = even and digit sum divisible by 3

Digit sum 0 + 1 + 2 + 3 + 4 + 5 = 15 is divisible by 3, so every arrangement is divisible by 3; the number need only be even.
Units digit 0: the other 5 digits fill the first five places in 5! = 120 ways.
Units digit 2 or 4: first digit is non-zero (4 choices), the remaining 4 places in 4! = 24 ways, giving 2 × 4 × 24 = 192.
Total = 120 + 192 = 312.

Answer: (b) 25%

Method. favourable count over total count

Total 5-digit numbers: first digit non-zero (4 ways), then 4! = 24 for the rest, so 4 × 24 = 96.
Greater than 50,000: first digit must be 5 (50,000 itself needs repeated 0s), remaining digits in 4! = 24 ways.
Percentage = 2496 × 100 = 25%.

Answer: (a) 625

Method. multiplication principle, repetition allowed

Odd digits: 1, 3, 5, 7, 9 (5 choices), none of them 0, so every place can take any of them.
Count = 5 × 5 × 5 × 5 = 625.

Answer: (a) 32

Method. fill the restricted places first (hundreds and units)

Numbers between 300 and 400 have hundreds digit 3.
Odd units digit, different from 3: 1, 5, 7 or 9, so 4 ways.
Tens digit: any of the 10 digits except the two already used: 8 ways.
Count = 1 × 8 × 4 = 32.

Answer: (c) 90

Method. restrict the hundreds digit, then fill the remaining places from the rest

A 3-digit number from S is less than 500 only if its hundreds digit is 2, 3 or 4 → 3 choices.
The tens and units digits are then chosen from the 6 remaining digits of S, all different: 6 × 5 = 30.
Total = 3 × 30 = 90.

Answer: (a) 240

Method. one digit repeats; arrangements of a multiset

Five places hold four different digits, each appearing at least once, so exactly one digit appears twice.
Choose the repeated digit: 4 ways.
Arrange the five digits (one pair alike): 5!2! = 60 ways.
Number of numbers = 4 × 60 = 240 (no 0 among the digits, so every arrangement is a 5-digit number).

Answer: (d) 0

Method. divisibility test for 3: the digit sum is fixed

Every such number uses all five digits 1, 2, 3, 4, 5 once, so its digit sum is 1 + 2 + 3 + 4 + 5 = 15.
15 is divisible by 3, so every one of these numbers is divisible by 3 (and is greater than 3).
Hence none of them is prime: the number is 0.

Answer: (c) 24

Method. fill the restricted (units) place first

An even number ends in 2 or 4 → 2 choices for the units digit.
Hundreds and tens places are filled from the remaining 4 digits: 4 × 3 = 12.
Total = 2 × 12 = 24.

Answer: (b) 2481

Method. complement: count numbers with all digits distinct and subtract

The integers lying between 2999 and 8001 are 3000, 3001, …, 8000.
From 3000 to 7999 (5000 numbers): all four digits distinct in 5 × 9 × 8 × 7 = 2520 ways (first digit 3–⁠7, then 9, 8, 7 choices).
So 5000 − 2520 = 2480 of them have at least two equal digits.
8000 also has equal digits (three 0s), so x = 2480 + 1 = 2481.

Answer: (c) 27216

Method. fill places left to right; the first digit cannot be 0

First (ten-thousands) digit: any of 1–⁠9 → 9 choices.
The remaining four places take distinct digits from the 9 digits left (0 now allowed): 9 × 8 × 7 × 6.
Total = 9 × 9 × 8 × 7 × 6 = 27216.

Shortcut. All arrangements minus those starting with 0: ¹⁰P₅ − ⁹P₄ = 30240 − 3024 = 27216.

Answer: (a) 2664

Method. sum of all permutations of distinct digits: each digit appears equally often in each place

There are 3! = 6 numbers.
Each digit appears 2 times in each place, so each place contributes 2(3 + 4 + 5) = 24.
Sum = 24 × (100 + 10 + 1) = 24 × 111 = 2664.

Answer: (d) 60

Method. numbers between 100 and 1000 are the three-digit numbers; arrange 3 of 5 distinct digits

Every number between 100 and 1000 is a three-digit number.
No digit is 0, so any 3 of the 5 digits in any order give a valid number: ⁵P₃ = 5 × 4 × 3 = 60.

Answer: (a) 24

Method. divisible by 10 ⇔ last digit 0

A number is divisible by 10 only if its units digit is 0, so the units place is fixed as 0.
The first three places are filled from 1, 5, 6, 7 without repetition: 4 × 3 × 2 = 24.
(The leading digit is never 0, since 0 is already used.)
Number of such numbers = 24.

Answer: (d) 1332

Method. sum of all permutations of distinct digits: each digit appears equally often in each place

There are 3! = 6 numbers.
Each digit appears in each place (units, tens, hundreds) 63 = 2 times.
Sum of the digits in any one place = 2(1 + 2 + 3) = 12.
Total = 12 × (100 + 10 + 1) = 12 × 111 = 1332.

Shortcut. Sum = (n − 1)! × (sum of digits) × 111…1 = 2! × 6 × 111 = 1332.

Answer: (c) 288

Method. consonant block plus vowels, with A repeated

CAPITAL: consonants C, P, T, L (4, distinct); vowels A, I, A (3, with A twice).
Arrange the consonant block with A, I, A: 4 units with A repeated, 4!/⁠2! = 12 ways.
Arrange the consonants inside the block: 4! = 24 ways.
Total = 12 × 24 = 288.

Answer: (c) 12

Method. vowels permute among vowel places, consonants among consonant places

DELHI: vowels E, I occupy positions 2 and 5; consonants D, L, H occupy positions 1, 3, 4.
Vowels among their places: 2! = 2; consonants among theirs: 3! = 6.
Total = 2 × 6 = 12.

Answer: (b) 6

Method. place vowels in odd positions, consonants in the rest

INDIA has vowels I, I, A and consonants N, D; the odd positions are 1, 3, 5 (three of them).
Vowels I, I, A in the 3 odd places: 3!/⁠2! = 3 ways.
Consonants N, D in positions 2, 4: 2! = 2 ways.
Total = 3 × 2 = 6.

Answer: (c) 1440

Method. treat each group as one block

EQUATION: vowels E, U, A, I, O (5); consonants Q, T, N (3); all letters distinct.
The vowel block and the consonant block can be ordered in 2! ways.
Inside the blocks: 5! ways for vowels, 3! ways for consonants.
Total = 2! × 5! × 3! = 2 × 120 × 6 = 1440.

Answer: (c) x = 4y

Method. arrangements of a word with repeated letters: n!p! q! …

PERMUTATIONS: 12 letters, T twice, rest different, so x = 12!2!.
COMBINATIONS: 12 letters, O, I, N twice each, rest different, so y = 12!2! 2! 2! = 12!8.
xy = 82 = 4, so x = 4y.

Answer: (c) 907200

Method. dictionary order: only words beginning with A come before C

Letters of MATHEMATICS: M, A, T twice each; H, E, I, C, S once (11 letters).
In alphabetical order the distinct letters are A, C, E, H, I, M, S, T, so only words beginning with A precede the first word beginning with C.
Fix A first; the other 10 letters are M, M, T, T, A, H, E, I, C, S.
Number of such words = 10!2! 2! = 36288004 = 907200.

Answer: (d) 864

Method. select vowels and consonants, then arrange

QUESTION has 8 different letters: vowels U, E, I, O (4) and consonants Q, S, T, N (4).
Choose 2 vowels and 2 consonants: ⁴C₂ × ⁴C₂ = 6 × 6 = 36 ways.
Arrange the 4 different letters: 4! = 24 ways.
Words = 36 × 24 = 864.

Answer: (c) 1152

Method. alternate pattern: choose the pattern, then arrange each group

4 vowels and 4 consonants alternate in 8 places in two patterns: V C V C … or C V C V ….
In each pattern the vowels fill their 4 places in 4! ways and the consonants theirs in 4! ways.
Words = 2 × 4! × 4! = 2 × 24 × 24 = 1152.

Answer: (b) 2880

Method. treat the block of consonants as one unit

Tie the 4 consonants Q, S, T, N into one block.
Units to arrange: the block and the 4 vowels, i.e. 5 units: 5! = 120 ways.
Inside the block the consonants arrange in 4! = 24 ways.
Words = 120 × 24 = 2880.

Answer: (b) 36

Method. place the vowels in the allowed (odd) positions first

TIGER has 5 different letters; vowels I, E; positions 1–⁠5, even positions 2 and 4.
The two vowels must go in the odd positions 1, 3, 5: ³P₂ = 6 ways.
The three consonants T, G, R fill the remaining three places: 3! = 6 ways.
Permutations = 6 × 6 = 36.

Answer: (a) 240

Method. choose the letters, then arrange them

LUCKNOW has 7 different letters: vowels U, O (2) and consonants L, C, K, N, W (5).
A word with two vowels uses both U and O and 2 of the 5 consonants: ²C₂ × ⁵C₂ = 1 × 10 = 10 selections.
Each selection of 4 different letters can be arranged in 4! = 24 ways.
Number of words = 10 × 24 = 240.

Answer: (b) 4

Method. place the vowels in the even places, consonants in the odd places

ABLE has 4 letters: vowels A, E and consonants B, L.
Even places are the 2nd and 4th: the 2 vowels fill them in 2! = 2 ways.
The 2 consonants fill the 1st and 3rd places in 2! = 2 ways.
Number of arrangements = 2 × 2 = 4.

Answer: (a) 19958400

Method. permutations with repeated letters: n!p!

PERMUTATION has 11 letters: P, E, R, M, U, T, A, T, I, O, N; only T is repeated (twice).
Number of permutations = 11!2! = 399168002 = 19958400.

Answer: (c) 210 × 8! × 8!

Method. choose which of the free people sit on each side, then arrange each side

Side A must hold the 4 particular men, side B the 2 particular men; 16 − 6 = 10 people are free.
Side A has 8 − 4 = 4 vacant chairs: choose its 4 extra people from the 10 in ¹⁰C₄ = 210 ways; the other 6 go to side B (which has 8 − 2 = 6 vacant chairs).
The 8 people on each side are then arranged in 8! ways.
Number of ways = 210 × 8! × 8!.

Answer: (b) 4320

Method. fill the restricted end places first, then arrange the rest

EQUATION has 8 distinct letters: vowels E, U, A, I, O (5) and consonants Q, T, N (3).
First place: 3 consonants to choose from; last place: 2 remaining consonants → 3 × 2 = 6 ways.
The other 6 letters fill the 6 middle places in 6! = 720 ways.
Number of words = 6 × 720 = 4320.

Answer: (c) 4

Method. ratio ⁿC_r+1/⁠ⁿC_r = (n − r)/⁠(r + 1)

¹⁵C_r+1/⁠¹⁵C_r = (15 − r)/⁠(r + 1) > 2.
15 − r > 2r + 2, so 3r < 13, r < 13/⁠3, hence r ≤ 4.
Check: r = 4: ¹⁵C₅ = 3003 > 2 × 1365 = 2730 (holds); r = 5: ¹⁵C₆ = 5005 < 2 × 3003 = 6006.
Greatest r = 4.

Answer: (c) 3 × ^m+1C₄

Method. expand both sides as polynomials in m

n = m(m − 1)/⁠2, so ⁿC₂ = n(n −⁠ 1)/⁠2 = (1/⁠2)·(m(m − 1)/⁠2)·((m² - m − 2)/⁠2) = m(m − 1)(m − 2)(m + 1)/⁠8.
^m+1C₄ = (m + 1)m(m − 1)(m − 2)/⁠24.
So ⁿC₂ = 3 × ^m+1C₄.

Shortcut. Put m = 5: n = 10, ¹⁰C₂ = 45; ⁶C₄ = 15, so the factor is 3 (option (c)); ⁷C₄ = 35 rules out (d). (At m = 4, (d) also gives 15, so m = 4 alone does not decide.)

Answer: (b) 2n

Method. ⁿC_a = ⁿC_b with a ≠ b gives a + b = n

⁵ⁿC_r = ⁵ⁿC_n+r, and r ≠ n + r (n ≥ 1).
So r + (n + r) = 5n, giving 2r = 4n, r = 2n.

Answer: (c) Both 1 and 2

Method. greatest binomial coefficient in a row of Pascal's triangle

C(n, r) increases while r < n2 and decreases after it.
If n is even, the greatest is at r = n2, i.e. n = 2r: statement 1 is correct.
If n is odd, the greatest is at r = n − 12 and r = n + 12, i.e. at the r with n = 2r + 1 and the r with n = 2r − 1 (the two values are equal): statement 2 is correct.
E.g. n = 7: C(7, 3) = C(7, 4) = 35 is the maximum. Both statements are correct.

Answer: (d) 21

Method. ⁿC_x = ⁿC_y ⇒ x = y or x + y = n

C(3n, 2n) = C(3n, 2n −⁠ 7) with 2n ≠ 2n − 7, so 2n + (2n − 7) = 3n.
4n − 7 = 3n ⇒ n = 7.
C(n, n −⁠ 5) = C(7, 2) = 7 × 62 = 21.

Answer: (c) C(51, 51) − C(51, 0)

Method. symmetry C(n, r) = C(n, n −⁠ r) cancels the terms in pairs

Since 21 + 30 = 22 + 29 = 23 + 28 = 24 + 27 = 25 + 26 = 51, C(51, 21) = C(51, 30), C(51, 22) = C(51, 29), C(51, 23) = C(51, 28), C(51, 24) = C(51, 27), C(51, 25) = C(51, 26).
Group: [C(51, 21) − C(51, 30)] + [C(51, 29) − C(51, 22)] + [C(51, 23) − C(51, 28)] + [C(51, 27) − C(51, 24)] + [C(51, 25) − C(51, 26)] = 0.
And C(51, 51) − C(51, 0) = 1 − 1 = 0, so option (c) equals the sum.

Answer: (a) 7

Method. AP condition 2b = a + c with ratios of consecutive binomial coefficients

2C(n, 5) = C(n, 4) + C(n, 6). Divide by C(n, 4), using C(n, 5)C(n, 4) = n − 45 and C(n, 6)C(n, 4) = (n − 4)(n − 5)30:
2(n − 4)5 = 1 + (n − 4)(n − 5)30.
Multiply by 30: 12(n − 4) = 30 + (n − 4)(n − 5) ⇒ n² − 21n + 98 = 0 ⇒ (n − 7)(n − 14) = 0.
n = 7 or 14; of the options only 7 fits (check: C(7, 4), C(7, 5), C(7, 6) = 35, 21, 7, common difference −⁠14).

Answer: (c) 10

Method. ⁿC_a = ⁿC_b with a ≠ b ⇒ a + b = n

n + 2 ≠ n − 2, so the equality needs (n + 2) + (n − 2) = 20.
2n = 20 ⇒ n = 10.
Check: C(20, 12) = C(20, 8).

Answer: (c) 28

Method. P(n, r) = r! · C(n, r) gives r, then n

P(n, r) = r! · C(n, r) ⇒ r! = 252021 = 120 ⇒ r = 5.
C(n, 5) = 21: C(7, 5) = C(7, 2) = 21, so n = 7 (check P(7, 5) = 7 × 6 × 5 × 4 × 3 = 2520).
C(n + 1, r + 1) = C(8, 6) = C(8, 2) = 28.

Answer: (c) C(52, 4)

Method. Pascal's rule C(m, 4) + C(m, 3) = C(m + 1, 4), applied in a chain

Combine from the smallest terms upward:
C(47, 4) + C(47, 3) = C(48, 4); + C(48, 3) = C(49, 4); + C(49, 3) = C(50, 4); + C(50, 3) = C(51, 4); + C(51, 3) = C(52, 4).
The sum is C(52, 4).

Answer: (b) 10

Method. ⁿC_a = ⁿC_b with a ≠ b ⇒ a + b = n

n + 2 ≠ n − 2, so the equality needs (n + 2) + (n − 2) = 20.
2n = 20 ⇒ n = 10.
Check: C(20, 12) = C(20, 8).

Answer: (d) C(n + 2, r)

Method. Pascal's rule C(n, r) + C(n, r −⁠ 1) = C(n + 1, r), used twice

Split the middle term: [C(n, r) + C(n, r −⁠ 1)] + [C(n, r −⁠ 1) + C(n, r −⁠ 2)].
By Pascal's rule this is C(n + 1, r) + C(n + 1, r −⁠ 1).
By Pascal's rule again it equals C(n + 2, r).

Answer: (c) 13

Method. distribute the prime factors of 210 = 2.3.5.7 among x, y, z

Ordered triples (x, y, z) of positive integers with xyz = 210: each of the 4 primes goes to one of 3 factors, 3⁴ = 81.
Two equal factors force x = y = 1 (210 is square-free): the 3 orderings of (1, 1, 210); remove them, 78 left with all entries distinct.
Each set of three distinct factors appears in 3! = 6 orders, exactly one of them with x < y < z: 78/⁠6 = 13.
(They are (1,2,105), (1,3,70), (1,5,42), (1,6,35), (1,7,30), (1,10,21), (1,14,15), (2,3,35), (2,5,21), (2,7,15), (3,5,14), (3,7,10), (5,6,7).)

No correct option. This question is defective as printed in UPSC's paper.

Method. sum of ⁶C_k for k = 0 to 3

'At most 3' means 0, 1, 2 or 3 things are selected.
Number = ⁶C₀ + ⁶C₁ + ⁶C₂ + ⁶C₃ = 1 + 6 + 15 + 20 = 42.

Defective as printed: the question does not say whether selecting nothing counts as a selection, and the two readings give two printed options. Counting the empty selection, C(6, 0) + C(6, 1) + C(6, 2) + C(6, 3) = 1 + 6 + 15 + 20 = 42, option (d); taking a selection to contain at least one thing, 6 + 15 + 20 = 41, option (c).

Answer: (c) 6

Method. stars and bars: positive solutions of x₁ + … + x_r = n number ^n-1C_r-1

Number of positive integer solutions of x + y + z = 5 is ^5-1C_3-1 = ⁴C₂ = 6.
(They are the arrangements of 1+1+3 (3 ways) and 1+2+2 (3 ways).)

Answer: (d) (n − 2)(n − 3)/⁠2

Method. fix the compulsory courses, choose the rest

The 2 compulsory courses are always taken, so the student chooses the remaining (n − 4) courses from the other (n − 2).
Ways = ^n-2C_n-4 = ^n-2C₂ = (n − 2)(n − 3)/⁠2.

Answer: (a) 20

Method. a strictly increasing triple is a 3-subset

Any 3 distinct faces from {1, …, 6} can be arranged in increasing order in exactly one way, giving x < y < z.
Number of outcomes = ⁶C₃ = 20.

Answer: (c) 485

Method. case split on how many women come from the man's side

Let the man invite k women and (3 − k) men; then the wife must invite (3 − k) women and k men, so that there are 3 women and 3 men in all.
Ways for a given k = (⁴C_k × ³C_3-k) × (³C_3-k × ⁴C_k) = (⁴C_k × ³C_3-k)².
k = 0: (1 × 1)² = 1; k = 1: (4 × 3)² = 144; k = 2: (6 × 3)² = 324; k = 3: (4 × 1)² = 16.
Total = 1 + 144 + 324 + 16 = 485.

Answer: (c) 64

Method. complement: total selections minus those with no black ball

Treat the 9 balls as distinct.
All selections of 3 = ⁹C₃ = 84.
Selections with no black ball (from 2 white + 4 red = 6) = ⁶C₃ = 20.
At least one black = 84 − 20 = 64.

Answer: (b) 21

Method. exclude the two players, choose from the rest

Leaving out the two particular players leaves 7 players.
Teams of 5 = ⁷C₅ = ⁷C₂ = 21.

Answer: (c) 21

Method. exclude the player, then choose from the rest

Leave out the particular player: 7 players remain.
Choose all 5 from these 7: ⁷C₅ = ⁷C₂ = 21.

Answer: (c) 246

Method. at least one = total minus none

All selections of 5 from 10 people: ¹⁰C₅ = 252.
Selections with no typist (all 5 from the 6 programmers): ⁶C₅ = 6.
Ways with at least one typist = 252 − 6 = 246.

Answer: (d) C(5, 3) × C(12, 8)

Method. select from each group separately and multiply

Bowlers: 5; non-bowlers: 17 − 5 = 12.
The team has 3 bowlers, chosen in C(5, 3) ways.
The other 11 − 3 = 8 players come from the 12 non-bowlers in C(12, 8) ways.
Number of teams = C(5, 3) × C(12, 8) (= 10 × 495 = 4950).

Answer: (d) 20

Method. a strictly decreasing triple corresponds to a 3-element subset

x > y > z means the three numbers are distinct.
Every choice of 3 different numbers from {1, …, 6} can be ordered in exactly one way as x > y > z.
Number of outcomes = ⁶C₃ = 20.

Answer: (c) 165

Method. count subsets of a given size by ⁿC_r

A has 10 elements.
Subsets with 2 elements: ¹⁰C₂ = 45.
Subsets with 3 elements: ¹⁰C₃ = 10.9.86 = 120.
Total = 45 + 120 = 165.

Answer: (c) 20

Method. two circles meet in at most 2 points

Each pair of circles gives at most 2 points: ⁵C₂ × 2 = 10 × 2 = 20.

Answer: (d) 52

Method. all triples minus collinear triples

Any 3 of the 8 points form a triangle unless all 3 come from the 4 collinear points (the other points are taken in general position).
Number of triangles = ⁸C₃ - ⁴C₃ = 56 − 4 = 52.

Answer: (c) 53

Method. reject 4-point sets containing 3 collinear points

A set of 4 points gives a quadrilateral only if no 3 of them are collinear.
Bad sets: 3 of the collinear points with 1 of the other 4: ⁴C₃ × 4 = 16; all 4 collinear points: 1.
Number = ⁸C₄ - 16 − 1 = 70 − 17 = 53.

Shortcut. Good sets have at most 2 collinear points: ⁴C₄ + ⁴C₁·⁴C₃ + ⁴C₂·⁴C₂ = 1 + 16 + 36 = 53.

Answer: (c) 8

Method. number of diagonals = n(n −⁠ 3)/⁠2

n(n −⁠ 3)/⁠2 = 20 gives n² - 3n − 40 = 0, i.e. (n − 8)(n + 5) = 0.
Since n > 0, n = 8.

Answer: (c) 14

Method. count line-line and line-circle intersections separately

Two lines meet in at most 1 point: ⁴C₂ × 1 = 6 points.
A line meets a circle in at most 2 points: 4 × 2 = 8 points.
Maximum total (all points distinct) = 6 + 8 = 14.

Answer: (a) 205

Method. all triples minus collinear triples

There are 3 + 4 + 5 = 12 points, and three points form a triangle unless all three lie on the same side.
Number of triangles = ¹²C₃ - ³C₃ - ⁴C₃ - ⁵C₃ = 220 − 1 − 4 − 10 = 205.

Answer: (c) 90

Method. two circles meet in at most 2 points

Each pair of circles gives at most 2 points, and for the maximum all these points are different.
Pairs of circles = ¹⁰C₂ = 45.
Maximum points = 2 × 45 = 90.

Answer: (d) 9

Method. a parallelogram is fixed by 2 lines from each family

Number of parallelograms = ^mC₂ × ⁿC₂ = 60.
^mC₂ and ⁿC₂ are triangular numbers 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …
The only factorisation of 60 into two such numbers is 6 × 10, so {^mC₂, ⁿC₂} = {6, 10}, giving {m, n} = {4, 5}.
m + n = 9.

Answer: (b) 2 only

Method. lines = ⁿC₂, triangles = ⁿC₃ for points with no three collinear

No three points on a circle are collinear, so every pair gives a different line and every triple gives a triangle.
Lines = ²⁰C₂ = 20 × 192 = 190, not 380, so statement 1 is wrong (380 counts each line twice).
Triangles = ²⁰C₃ = 20 × 19 × 186 = 1140, so statement 2 is correct.
Answer: 2 only.

Answer: (b) 50

Method. triangles with no side of the polygon: total minus those sharing one or two sides

Total triangles from the 10 vertices = ¹⁰C₃ = 120.
Triangles with two sides of the polygon (three consecutive vertices): one for each vertex in the middle = 10.
Triangles with exactly one side of the polygon: choose the side (10 ways), then the third vertex must avoid the two ends and their two neighbours: 10 − 4 = 6 ways, giving 10 × 6 = 60.
Triangles with no common side = 120 − 10 − 60 = 50.

Shortcut. Use n(n −⁠ 4)(n − 5)6 = 10 × 6 × 56 = 50.

No correct option. This question is defective as printed in UPSC's paper.

Method. two distinct circles meet in at most 2 points; each pair counted once

Intended count: two distinct circles cut in at most 2 points, and 5 circles give ⁵C₂ = 10 pairs, so at most 2 × 10 = 20 points — but every pair cutting in 2 points means every pair of circles overlaps.
Taken as printed, "non-overlapping" circles have disjoint insides, so two of them can meet only by touching, in at most 1 point.
All 10 pairs cannot touch at once (5 mutually touching circles with disjoint insides do not exist: their contact graph would be the non-planar K₅); at most 3 × 5 − 6 = 9 pairs can touch, e.g. three mutually touching circles, a fourth in their gap touching all three, a fifth in a gap of the fourth touching it and two of the first three.
So the printed question gives a maximum of 9, which is not an option. (If nested, internally touching circles are allowed, the same bound holds: only two circles touch three mutually touching circles, and those two do not touch each other.)

"Non-overlapping" circles cannot cut each other, so they meet only by touching and the maximum is 9 (no option, and option a's 10 is unattainable since 5 such circles cannot all touch pairwise); read as "no two circles coincident", the count is 2 × ⁵C₂ = 20, option c.

Answer: (d) 20

Method. diagonals = lines joining vertices minus sides

Any 2 of the 8 vertices give a segment: ⁸C₂ = 28.
8 of these segments are sides.
Number of diagonals = 28 − 8 = 20.

Shortcut. n(n −⁠ 3)2 = 8 × 52 = 20.

Answer: (b) 45

Method. a line is fixed by any 2 points when no three are collinear

No three points are collinear, so each pair of points gives a different line.
Number of lines = ¹⁰C₂ = 10 × 92 = 45.

Answer: (a) 185

Method. triangles = all triples minus collinear triples

Any 3 of the 12 points: ¹²C₃ = 220 triples.
A triple chosen from the 7 collinear points gives no triangle: ⁷C₃ = 35 such triples.
Number of triangles = 220 − 35 = 185.
(This assumes, as intended, that no other three of the points are collinear and none of the other five lies on that line.)