Permutations And Combination Model Questions & Answers, Practice Test for ibps po mains 2024 2025

ibps po mains 2024 2025 SYLLABUS WISE SUBJECTS MCQs

Number System

Simplification

Ratio & Proportion

Percentage

Time & Work

Time & Distance

Mensuration

Data Sufficiency

Average

Profit & Loss

Alligation & Mixtures

Compound Interest

Number Series

Linear Equations

Quadratic Equations

Probability

Permutations & Combination

Click To More Exercise For Permutations & Combination
Question :1

If a team of four persons is to be selected from 8 males and 8 females, then in how many ways can the selections be made to include at least one male.

Answer: (d)

1 m + 3f = $^8C_1 × ^8C_3$ = 8 × 56 = 448

2 m + 2f = $^8C_2 × ^8C_2$ = 28 × 28 = 784

3 m + 1f = $^8C_3 × ^8C_1$ = 56 × 8 = 448

4 m + 8f = $^8C_4 × ^8C_0$ = 70 × 1 = 70

Total = 1750

Question :2

The total number of ways in which letters of the word ACCOST can be arranged so that the two C's never come together will be

Answer: (a)

Total number of ways to permute 6 alphabets 2 of which are common = 6! / 2! = 360.

(1) Treat the two C's as one

⇒ Number of possible ways = $^5P_5$ = 120

(b) Number of ways = Total arrangements – Number of arrangements in which they always come together

= 360 – 120 = 240.

Question :3

If P(32, 6) = kC (32, 6), then what is the value of k?

Answer: (d)

Since $^32P_6 = k ^32C_6$

⇒ ${32!}/(32 - 6) = k . {32!}/{6!(32 6)!}$

⇒ k = 6! = 720

Question :4

A man has 3 shirts, 4 trousers and 6 ties. What are the number of ways in which he can dress himself with a combination of all the three?

Answer: (b)

$^3C_1 × $4C_1 × ^6C_1$ = 72

Question :5

In how many ways can 13 different alphabets (a, b, c, ... m) be arranged so that the alphabets f and g never come together?

Answer: (a)

Total possible arrangements = $^13P_13$ = 13!

Total number in which f and g are together = 2 × $^12P_12$ = 2 × 12!

Click To More Exercise For Permutations & Combination
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