68 free MCQs on Permutations and Combinations with worked answers and explanations. Counting, arrangements, and selections
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Counting tree for selecting 2 items from {A, B, C} without repetition: 3 x 2 = 6 ordered arrangements.
Q1.
5! (5 factorial) =
Answer: C. 120
Why: 5! = 5×4×3×2×1 = 120.
Q2.
0! =
Answer: B. 1
Why: By convention, 0! = 1. This makes combinatorial formulas work correctly.
Q3.
nPr represents:
Answer: B. Number of arrangements of r from n distinct objects
Why: nPr = n!/(n-r)! counts ordered arrangements (permutations) of r items from n distinct items.
Q4.
nCr represents:
Answer: B. Number of ways to select r items from n (order does not matter)
Why: nCr = n!/[r!(n-r)!] counts unordered selections (combinations).
Q5.
5P2 =
Answer: B. 20
Why: 5P2 = 5!/(5-2)! = 5!/3! = 5×4 = 20.
Q6.
5C2 =
Answer: A. 10
Why: 5C2 = 5!/(2! × 3!) = (5×4)/(2×1) = 10.
Q7.
nC<sub>0</sub> =
Answer: C. 1
Why: nC<sub>0</sub> = n!/(0! × n!) = 1. There is exactly one way to choose nothing.
Q8.
nCn =
Answer: B. 1
Why: nCn = 1. There is exactly one way to choose all items.
Q9.
The number of ways to arrange 4 books on a shelf:
Answer: C. 24
Why: 4P4 = 4! = 24. All 4 books can be arranged in 4! = 24 ways.
Q10.
How many ways can 3 people sit in a row of 5 seats?
Answer: C. 60
Why: 5P3 = 5!/(5-3)! = 5×4×3 = 60.
Q11.
nCr = nC(n-r). So 10C3 = 10C?
Answer: A. 7
Why: 10C3 = 10C(10-3) = 10C7. So 10C3 = 10C7.
Q12.
How many ways can we choose 2 out of 6 flavors of ice cream?
Answer: B. 15
Why: 6C2 = 15. Order does not matter (just choosing, not arranging).
Q13.
The number of subsets of a set with 4 elements:
Answer: C. 16
Why: Total subsets = 2<sup>n</sup> = 2<sup>4</sup> = 16 (including empty set and the set itself).
Q14.
nPr = r! × nCr. True or false?
Answer: A. True
Why: nPr = n!/(n-r)! and nCr = n!/[r!(n-r)!]. So nPr = r! × nCr. True.
Q15.
How many 2-digit numbers can be formed from digits 1,2,3,4,5 without repetition?
Answer: B. 20
Why: 5P2 = 5×4 = 20 arrangements (order matters, different 2-digit numbers).
Q16.
In how many ways can a committee of 3 be chosen from 7?
Answer: B. 35
Why: 7C3 = 35. Order does not matter in a committee.
Q17.
Multiplication principle: if 3 choices for first and 4 for second, total combinations:
Answer: B. 3×4=12
Why: Multiplication principle: 3 × 4 = 12. When choices are independent and both must be made.
Q18.
How many 3-letter words from ABCDE (letters not repeated)?
Answer: C. 60
Why: 5P3 = 5×4×3 = 60.
Q19.
Pascal triangle entry in row n, column r is:
Answer: B. nCr
Why: Pascal triangle entries are binomial coefficients nCr. Row n (starting from 0) has entries nC<sub>0</sub>, nC<sub>1</sub>,...,nCn.
Q20.
A coin is tossed 3 times. Number of ways to get exactly 2 heads:
Answer: B. 3
Why: 3C2 = 3. The three possible outcomes: HHT, HTH, THH.
Q21.
In how many ways can 5 people be seated in a circle?
Answer: B. 4!
Why: Circular permutations of n objects = (n-1)! = 4! = 24.
Q22.
Number of ways to arrange letters in MISSISSIPPI:
Answer: B. 11!/(4!4!2!)
Why: MISSISSIPPI: 11 letters. M=1, I=4, S=4, P=2. Arrangements = 11!/(1!4!4!2!).
Q23.
Number of ways to select at least one from n objects:
Answer: B. 2<sup>n</sup> - 1
Why: Total subsets = 2<sup>n.</sup> Excluding empty set: 2<sup>n</sup> - 1 ways to select at least one object.
Q24.
Using binomial theorem, coefficient of x³ in (1+x)<sup>7</sup> is:
Answer: B. 35
Why: 7C3 = 35. Coefficient of x<sup>r</sup> in (1+x)<sup>n</sup> is nCr.
Q25.
Sum of all coefficients in (x+y)<sup>n</sup> is:
Answer: A. 2<sup>n</sup>
Why: Set x=y=1: (1+1)<sup>n</sup> = 2<sup>n.</sup> Sum of all binomial coefficients = 2<sup>n.</sup>
Q26.
nC(r-1) + nCr = ?
Answer: B. (n+1)Cr
Why: Pascal identity: nC(r-1) + nCr = (n+1)Cr. Basis of Pascal triangle.
Q27.
How many 4-digit numbers can be formed from 1-9 without repetition?
Answer: C. 9P4
Why: 9P4 = 9×8×7×6 = 3024. Options A and C are equivalent.
Q28.
In how many ways can a team of 3 boys and 2 girls be chosen from 5 boys and 4 girls?
Answer: B. 5C3 × 4C2 = 60
Why: 5C3 × 4C2 = 10 × 6 = 60. Independent choices multiplied together.
Q29.
Number of diagonals in a hexagon (6-sided polygon):
Answer: B. 9
Why: Diagonals = nC<sub>2</sub> - n = n(n-1)/2 - n = n(n-3)/2. For n=6: 6×3/2 = 9.
Q30.
The number of triangles formed by 10 points (no 3 collinear):
Answer: B. 120
Why: Any 3 points form a triangle. 10C3 = 120.
Q31.
How many words can be formed from GARDEN using all letters?
Answer: B. 720
Why: GARDEN: 6 distinct letters. 6! = 720 arrangements.
Q32.
ABCD... if A must be first: arrangements of 5 people ABCDE with A first:
Answer: A. 4!
Why: A fixed in position 1. Remaining 4 can be arranged in 4! = 24 ways.
Q33.
Number of ways to distribute 5 identical balls into 3 distinct boxes:
Answer: B. 21
Why: Stars and bars: C(5+3-1, 3-1) = C(7,2) = 21.
Q34.
The number of permutations of n things taken all at once when p are alike:
Answer: A. n!/p!
Why: When p items are identical: permutations = n!/p!. Divides by repetitions.
Q35.
How many ways can 10 people be divided into two groups of 5?
Answer: B. 10C5 / 2 = 126
Why: 10C5 = 252. But groups are unordered (not labeled), so divide by 2: 252/2 = 126.
Q36.
Words from BOOK (all permutations):
Answer: C. 12 (O repeated twice)
Why: BOOK: 4 letters with O repeated twice. 4!/2! = 12.
Q37.
Total number of functions from set A (3 elements) to set B (4 elements):
Answer: B. 64
Why: Each element of A can map to any of 4 elements in B. Total = 4<sup>3</sup> = 64.
Q38.
Number of injective (one-to-one) functions from A (3 elements) to B (4 elements):
Answer: D. 4P3 = 24
Why: Injective: no two elements of A map to same element in B. 4P3 = 4×3×2 = 24.
Q39.
In how many ways can 8 people be arranged in 2 rows of 4?
Answer: A. 8! = 40320
Why: All 8 in 2 rows of 4 is still 8! since specific positions matter. But if rows are 4 fixed positions each: 8P8 = 8! = 40320.
Q40.
The middle term in the expansion of (1+x)<sup>2n</sup> is:
Answer: A. (n+1)th term with coefficient 2nCn
Why: Expansion of (1+x)<sup>2n</sup> has 2n+1 terms. Middle term is (n+1)th term: C(2n,n) × x<sup>n.</sup>
Q41.
The number of ways to distribute n identical objects into r distinct groups (some may be empty):
Answer: B. C(n+r-1, r-1)
Why: Stars and bars: place n stars and r−1 bars. Total positions = n+r−1, choose r−1 for bars. Formula: C(n+r−1, r−1). Allows empty groups.
Q42.
Number of non-negative integer solutions to x₁ + x₂ + x₃ = 10:
Answer: A. 66
Why: Stars and bars with n=10, r=3 groups. Solutions = C(10+3−1, 3−1) = C(12,2) = 12×11/2 = 66.
Q43.
Derangement D(n): number of permutations where no element is in original position. D(4) =
Answer: A. 9
Why: Inclusion-exclusion: D(4) = 4!(1−1+1/2!−1/3!+1/4!) = 24(1/2−1/6+1/24) = 24(12/24−4/24+1/24) = 24×9/24 = 9.
Q44.
Total number of permutations of n items with repetitions: n₁ alike, n₂ alike,..., nk alike (sum = n) =
Answer: B. n!/(n₁! n₂! ... nk!)
Why: Multinomial coefficient: start with n! total arrangements, divide by n₁! for identical group 1, n₂! for group 2, etc. Result: n!/(n₁!n₂!…nk!).
Q45.
Number of ways to seat 5 couples in a row such that each couple sits together:
Answer: A. 5! × 2⁵
Why: Block method: treat each couple as 1 unit → 5 blocks arranged in 5! = 120 ways. Each couple can swap internally: 2 ways each → 2⁵ = 32. Total = 120×32 = 3840 = 5!×2⁵.
Q46.
In how many ways can 4 red, 3 blue, and 2 green balls be arranged in a row?
Answer: B. 9!/(4!3!2!)
Why: Multinomial permutation: n=9 total, with groups 4,3,2. Arrangements = 9!/(4!3!2!) = 362880/(24×6×2) = 362880/288 = 1260.
Q47.
Number of ways to select 3 from 6 where order matters for first 2 but not the third:
Answer: C. 6P2 × 4C1
Why: Step 1: arrange first 2 from 6 in order → 6P2 = 6×5 = 30. Step 2: pick the 3rd from remaining 4 (unordered) → 4C1 = 4. Total = 30×4 = 120 = 6P2×4C1.
Q48.
Catalan number Cₙ = C(2n,n)/(n+1). C₃ =
Answer: A. 5
Why: C₃ = C(6,3)/(3+1) = 20/4 = 5. Verify: C(6,3) = 6!/(3!3!) = 720/36 = 20. Catalan numbers count valid bracket sequences, triangulations, etc.
Q49.
The number of lattice paths from (0,0) to (m,n) moving only right or up:
Answer: B. C(m+n, m)
Why: Each path needs exactly m right-steps (R) and n up-steps (U): total m+n steps. Choose which m of the m+n steps are R: C(m+n, m). Remaining n are automatically U.
Q50.
Number of binary strings of length n with exactly k ones:
Answer: B. C(n,k)
Why: A binary string of length n with exactly k ones: choose k positions (out of n) to place the 1s; the rest become 0s. Number of ways = C(n,k).
Q51.
Inclusion-Exclusion: |A ∪ B ∪ C| =
Answer: B. |A|+|B|+|C|-|A∩B|-|A∩C|-|B∩C|+|A∩B∩C|
Why: Add singles, subtract pairwise intersections (over-subtracted), add back triple (under-subtracted). |A∪B∪C| = |A|+|B|+|C| − |A∩B| − |A∩C| − |B∩C| + |A∩B∩C|.
Q52.
How many 5-digit palindromes exist? (Form abcba)
Answer: A. 900
Why: Form abcba: 'a' determines 1st & 5th, 'b' 2nd & 4th, 'c' middle. a ∈ {1–9}: 9 choices; b ∈ {0–9}: 10; c ∈ {0–9}: 10. Total = 9×10×10 = 900.
Q53.
Number of ways to arrange n people in a circle with one fixed (labeled) position:
Answer: B. (n-1)!
Why: Circular permutation: fix one person to remove rotational equivalence. The remaining n−1 people arrange linearly in (n−1)! ways. Answer: (n−1)!.
Q54.
In a group of 10 people, handshakes if everyone shakes everyone else exactly once:
Answer: B. 45
Why: Each handshake = 1 unique pair of people. Count pairs from 10: C(10,2) = 10×9/2 = 45. Order within a pair doesn't matter, so no doubling.
Q55.
Number of positive divisors of 360 = 2³ × 3² × 5:
Answer: C. 24
Why: 360 = 2³ × 3² × 5¹. Each divisor chooses exponent: 0–3 for 2 (4 choices), 0–2 for 3 (3 choices), 0–1 for 5 (2 choices). Total = 4×3×2 = 24.
Q56.
Total number of onto functions from A (m elements) to B (n elements) using inclusion-exclusion:
Answer: B. sum from k=0 to n of (-1)<sup>k</sup> × C(n,k) × (n-k)<sup>m</sup>
Why: Inclusion-exclusion over elements of B left uncovered: Σₖ₌₀ⁿ (−1)ᵏ C(n,k)(n−k)ᵐ. Each term accounts for functions missing exactly k elements of B.
Q57.
The Stirling number S(n,k) counts:
Answer: B. Partitions of n-element set into k non-empty subsets
Why: Stirling numbers of the 2nd kind S(n,k): partition a set of n distinct elements into exactly k non-empty, unordered subsets. E.g. S(3,2)=3: {1},{2,3} and permutations.
Q58.
Number of ways to choose a committee of 5 from 10 where at least 2 women (4 women, 6 men):
Answer: A. 186
Why: Exactly 2 women: C(4,2)×C(6,3) = 6×20 = 120. Exactly 3 women: C(4,3)×C(6,2) = 4×15 = 60. Exactly 4 women: C(4,4)×C(6,1) = 1×6 = 6. Total = 120+60+6 = 186.
Q59.
The generating function for combinations is related to:
Answer: A. (1+x)<sup>n</sup> = sum nCr × x<sup>r</sup>
Why: Binomial theorem: (1+x)ⁿ = Σᵣ₌₀ⁿ C(n,r)xʳ. The coefficient of xʳ is exactly C(n,r). This is the ordinary generating function for the sequence of binomial coefficients.
Q60.
How many ways to arrange letters AABBCC?
Answer: A. 90
Why: 6 letters with A repeated 2×, B repeated 2×, C repeated 2×. Multinomial: 6!/(2!×2!×2!) = 720/8 = 90. Each of 90 arrangements is distinct.
Q61.
The number of distinct arrangements of the letters of the word MISSISSIPPI is:
Answer: A. 34650
Why: 11 letters with I×4, S×4, P×2: 11!/(4!·4!·2!) = 34650.
Q62.
The number of 4-digit numbers with all distinct digits chosen from 1–9 that are even is:
Answer: A. 1344
Why: Units digit even: 4 choices. Remaining three places from the other 8 digits: 8·7·6 = 336. Total 4·336 = 1344.
Q63.
The number of diagonals of a convex 12-sided polygon is:
Answer: B. 54
Why: Diagonals = n(n−3)/2 = 12·9/2 = 54.
Q64.
The number of ways to seat 5 boys and 3 girls in a row so that no two girls are adjacent is:
Answer: B. 14400
Why: Arrange 5 boys: 5! = 120. Choose and fill 3 of the 6 gaps: 6·5·4 = 120. Total 120·120 = 14400.
Q65.
From 6 men and 4 women, the number of 5-member committees with at least 2 women is:
Answer: C. 186
Why: Total C(10,5) = 252, minus 0 women C(6,5) = 6 and exactly 1 woman C(4,1)C(6,4) = 60. So 252 − 66 = 186.
Q66.
The sum of all 4-digit numbers formed using each of 1, 2, 3, 4 exactly once is:
Answer: A. 66660
Why: Each digit occupies each place 3! = 6 times; digit sum = 10. Sum = 6·10·1111 = 66660.
Q67.
The number of ways to distribute 10 identical balls into 3 distinct boxes with each box non-empty is:
Answer: A. 36
Why: Stars and bars with each ≥ 1: C(10−1, 3−1) = C(9,2) = 36.
Q68.
The number of positive integers less than 1000 divisible by neither 5 nor 7 is:
Answer: B. 686
Why: Among 1–999: divisible by 5 is 199, by 7 is 142, by 35 is 28. By 5 or 7 = 313, so neither = 999 − 313 = 686.