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Aptitude & Reasoning · Question Set
Permutations, Combinations & Probability interview questions for placements and exams.
Questions
14
Included in this set
Subject
Aptitude & Reasoning
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Difficulty
Mixed
Level of this set
Go through each question and its explanation. Use this set as a focused practice pack for Aptitude & Reasoning.
Hearts are 13. Kings are 4. King of hearts is counted twice, subtract one. Total favorable is 16 plus 1 equals 17. Divide by 52.
(13 + 4 − 1)/52 = 17/52
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Pick order for three places from four distinct digits. That is 4 permute 3 equals twenty four.
P(4,3) = 4×3×2 = 24
For circular permutations with only rotation identical, use (n−1) factorial. For seven people, six factorial equals seven hundred twenty.
(7−1)! = 6! = 720
Even last digit choices: 0, 2, 4 gives three. First digit cannot be zero, so for last digit each case consider: If last is 0, first has 4 options (1–4). If last is 2 or 4, first has 4 options (0,1,3,4 or 0,1,2,3). Middle two each have 5 options. Count: Case last 0 → 4×5×5 = 100. Case last 2 or 4 → 2×(4×5×5) = 200. Total 300.
100 + 200 = 300
Use stars and bars for nonnegative integer solutions to x1+x2+x3+x4=10. The count is C(10+4−1, 4−1) equals C(13,3) equals two hundred eighty six.
C(13,3) = 286
Favorable outcomes are 5 and 6 which are two outcomes out of six. Two by six simplifies to one by three.
P = 2/6 = 1/3
First red is five eighths. Then four sevenths. Multiply to get twenty by fifty six which reduces to ten by twenty one.
(5/8)×(4/7) = 20/56 = 10/21
Use complement. Probability of no head is both tails which is one fourth. So at least one head is three fourths.
1 − (1/2×1/2) = 3/4
Seven letters with repeats: L twice, O twice. Divide by duplicates. Seven factorial over two factorial times two factorial equals five thousand forty over four equals one thousand two hundred sixty.
7!/(2!2!) = 5040/4 = 1260
Treat the two Cs as one block. We then arrange 6 items: block C C, A, C? Check letters: A C C O U N T has 7 letters with two C’s. Block makes 6 objects: [CC], A, O, U, N, T. These 6 are distinct, so 6 factorial equals seven hundred twenty.
6! = 720
Arrange n distinct items in n factorial ways. Six factorial equals seven hundred twenty.
6! = 720
When order does not matter, use combinations. Ten choose four equals two hundred ten.
C(10,4) = 210
Sum cases. 2 women: C(5,2)×C(8,3)=10×56=560. 3 women: C(5,3)×C(8,2)=10×28=280. 4 women: C(5,4)×C(8,1)=5×8=40. 5 women: C(5,5)×C(8,0)=1×1=1. Total 560+280+40+1=881? That seems off; we miscounted because we only have 5 women total, correct combinations are fine. Recompute: 560 + 280 + 40 + 1 = 881. None of the options match. We must have made an arithmetic slip earlier in option design. Fix counts: For 2 women: C(5,2)=10, C(8,3)=56 → 560. For 3 women: 10×28=280. For 4 women: 5×8=40. For 5 women: 1×1=1. Sum is 881. Provide correct option 881.
Σ = 560 + 280 + 40 + 1 = 881
Use Bayes. P(def) equals 0.6×0.02 plus 0.4×0.05 equals 0.012 plus 0.02 equals 0.032. P(B|def) equals (0.4×0.05)/0.032 equals 0.02/0.032 equals 0.625? Wait compute carefully: 0.4×0.05 = 0.02. Divide by 0.032 gives 0.625. So the correct answer is 0.625, not 0.4. Choose nearest? Instead provide the exact value.
P(B|D) = 0.02 / 0.032 = 0.625