# Permutation and Combination Quiz 11

Home > > Tutorial
5 Steps - 3 Clicks

# Permutation and Combination Quiz 11

### Introduction

Permutation and Combination is one of important topic in Quantitative Aptitude Section. In Permutation and Combination Quiz 11 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 11 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 numbers are divisible by 4?
A. 256 B. 225 C. 198 D. 252

B
A number is divisible by 4 when its last two digits are divisible by 4
For this the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96
By the formula, [latex]{a}_{n}[/latex] = a + (n - 1)d
96 = 0 + (n - 1) × 4
n = 25
So there are 25 choices for last 2 digits and 9 choices (1-9) for the 1st digit
So total 9 × 25 = 225

### Q2

There are 8 men and 7 women. In how many ways a group of 5 people can be made such that the particular woman is always to be included?
A. 860 B. 1262 C. 1768 D. 984

C
Total 15 people, and a particular woman is to be taken to form a group of 5, so choice is to be done from 14 people of 4 people
Ways are [latex]^{14}{C}_{4}[/latex] = 1768

### Q3

How many 4 digit words can be made from the digits 7, 8, 5, 0, and 4 without repetition?
A. 70 B. 96 C. 84 D. 104

B
0 cannot be on first place for it to be a 4 digit number.
So for 1st digit 4 choices, for second also 4 (because 0 can be placed here), then 3 for third place, 2 for fourth place
Total numbers = 4 × 4 × 3 × 2 = 96

### Q4

In how many ways 8 students can be given 3 prizes such that no student receives more than 1 prize?
A. 348 B. 284 C. 224 D. 336

D
For [latex]{1}^{st}[/latex] prize there are 8 choices, for [latex]{2}^{nd}[/latex] prize, 7 choices, and for [latex]{3}^{rd}[/latex] prize – 6 choices left
So total ways = 8 × 7 × 6 = 336

### Q5

A bag contains 4 red balls and 5 black balls. In how many ways can i make a selection so as to take atleast 1 red ball and 1 black ball ?
A. 564 B. 345 C. 465 D. 240

C
[latex]{2}^{4} - 1[/latex] = 16 - 1 = 15
[latex]{2}^{5} - 1[/latex] = 32 - 1 = 31
15 × 31 = 465

### Exams

Competitive Exams - Entrance Exams
UG NSTSE 2020 RIMC Admission 2020 WBJEE EVETS 2019
Diploma HPBOSE D.El.Ed CET 2019 Goa Diploma Admissions 2019
###### XAT 2020

Competitive Exams - Recent Job Notifications
Category
Banking SSC Railway
Defence Police Insurance