# Permutation and Combination Quiz 8

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# Permutation and Combination Quiz 8

### Introduction

Permutation and Combination is one of important topic in Quantitative Aptitude Section. In Permutation and Combination Quiz 8 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Permutation and Combination – Quiz 8 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

How many 3 digit number can be formed with the digits 5, 6, 2, 3, 7 and 9 which are divisible by 5 and none of its digit is repeated?
A. 12 B. 16 C. 20 D. 24

C
_ _ 5
first two places can be filled in 5 and 4 ways respectively so, total number of 3 digit number = 5 × 4 × 1 = 20

### Q2

In how many different ways can the letter of the word ELEPHANT be arranged so that vowels always occur together?
A. 2060 B. 2160 C. 2260 D. 2360

B
Vowels = E, E and A. They can be arranged in $(\frac {3!}{2!})$ Ways
so total ways = 6!$\times (\frac {3!}{2!})$ = 2160

### Q3

There are 4 bananas, 7 apples and 6 mangoes in a fruit basket. In how many ways can a person make a selection of fruits from the basket.
A. 269 B. 280 C. 279 D. 256

C
Zero or more bananas can be selected in 4 + 1 = 5 ways (0 orange, 1 orange, 2 orange, 3 orange and 4 orange)
Similarly apples can be selected in 7 + 1 = 8 ways
And mangoes in 6 + 1 = 7 ways
So total number of ways = 5 × 8 × 7 = 280
But we included a case of 0 orange, 0 apple and 0 mangoes, so we have to subtract this, so 280 – 1 = 279 ways

### Q4

There are 15 points in a plane out of which 6 are collinear. Find the number of lines that can be formed from 15 points.
A. 105 B. 90 C. 91 D. 95

C
From 15 points number of lines formed = $^{15}{C}_{2}$
6 points are collinear, number of lines formed by these = $^{6}{C}_{2}$
So total lines = $^{15}{C}_{2} – ^{15}{C}_{2}$ + 1 = 91

### Q5

In how many ways 4 Indians, 5 Africans and 7 Japanese be seated in a row so that all person of same nationality sits together
A. 4! 5! 7! 3! B. 4! 5! 7! 5! C. 4! 6! 7! 3! D. Can’t be determined

A
4 Indians can be seated together in 4! Ways, similarly for Africans and Japanese in 5! and 7! respectively. So total ways = 4! 5! 7! 3!

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