# Permutation and Combination – Quiz 7

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# Permutation and Combination – Quiz 7

### Introduction

Permutation and Combination is one of important topic in Quantitative Aptitude Section. In Permutation and Combination – Quiz 7 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 7 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 integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?
A. 376 B. 375 C. 500 D. 673

A
The smallest number in the series is 1000, a 4-digit number.
The largest number in the series is 4000, the only 4-digit number to start with 4.
The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.
The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.
Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.
Including 4000, there will be 376 such numbers.

### Q2

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :
A. 601 B. 600 C. 603 D. 602

A
If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.
If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.
If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.
The required word SACHIN can be obtained after the 5 X 5! = 600 Ways i.e. SACHIN is the ${601}^{th}$ letter.

### Q3

12 people at a party shake hands once with everyone else in the room. How many handshakes took place?
A. 72 B. 66 C. 76 D. 64

B
There are 12 people, so this is our n value.
So, $^{12}{C}_{2}$= 66

### Q4

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?
A. 1260 B. 1400 C. 1250 D. 1600

A
A team of 6 members has to be selected from the 10 players. This can be done in $^{10}{C}_{6}$ or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210 × 6 = 1260

### Q5

How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?
A. 216 B. 45360 C. 1260 D. 43200

D
There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.
The number of ways in which 9 letters can be arranged = $\frac {9!}{2! × 2! × 2!}$ = 45360
There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in $\frac {6!}{2!× 2!}$ = 180 ways.
In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in $\frac {4!}{2!}$ = 12 ways.
The number of ways in which the four vowels always come together = 180 x 12 = 2160.
Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

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