# Permutation and Combination Quiz 13

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# Permutation and Combination Quiz 13

### Introduction

Permutation and Combination is one of important topic in Quantitative Aptitude Section. In Permutation and Combination Quiz 13 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 13 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

Find the 25th word if the word JACKED is arranged in dictionary order.
A. ACDEJK B. ADCEJK C. AECDJK D. AEDCJK

B
As in JACKED total number of letters are 6 and the lowest letter as per dictionary is A so words starting with A are = 5! (number of letters left in JACKED other than A)
= 5 × 4 × 3 × 2 × 1 = 120 words so 25th word will definitely start with A
And after A next letter is C so words starting with AC are = 4!(number of letters left in JACKED other than A, C) = 4 × 3 × 2 × 1 = 24. So 25th word can’t start with AC.
And after AC in dictionary combination, to start words will be with AD and so 25th word will start with AD which is ADCEJK (letters other than AD are arranged in alphabetical order because dictionary follows this pattern)

### Q2

What will be the 55th word in the arrangements of the letters of the word PRANKED?

C
As in PRANKED total number of letters are 7 and the lowest letter as per dictionary is A so words starting with A are = 6! (number of letters left in PRANKED other than A)
= 6 × 5 × 4 × 3 × 2 × 1 = 720 words so 55th word will definitely start with A.
After A next letter is D so words starting with AD are = 5!(number of letters left in PRANKED other than AD) = 5 × 4 × 3 × 2 × 1 = 120 so 55th word will definitely start with AD.
After AD next letter is E so words starting with ADE are = 4!(number of letters left in PRANKED other than ADE) = 4 × 3 × 2 × 1 = 24 so 55th word is not included in it
And after combination of ADE next combination is ADK so words forming with ADK = 4!(number of letters left in PRANKED other than ADK) = 4 × 3 × 2 × 1 = 24 so 24 + 24 = 48 so 55th word is not included in it.
After combination of ADK next combination is ADN so words forming with ADN = 4!(number of letters left in PRANKED other than ADN) = 4 × 3 × 2 × 1 = 24 so 24 + 24 +24 = 72 so 55th word is included in it.
So ADNE will be the next combination and with it 3! = 3 × 2 × 1 = 6 words can form so 48 + 6 = 54
So after ADNE next combination is ADNK and with it 3! = 3 × 2 × 1 = 6 words can form so ADNKEPR is the 55th word.

### Q3

An organization has 3 committees, only 2 persons are members of all 3 committee but every pair of committee has 3 members in common. What is the least possible number of members on any one committee?
A. 4 B. 5 C. 6 D. 1

A
Say A, B and C are the three committees.
n(A ∩ B ∩ C) = 2, n(A ∩ B) = 3, n(A ∩ C) = 3 and n(B ∩ C) = 3
⇒ Number of people in committee A and B but not in C = 3 – 2 = 1
⇒ Number of people in committee A and C but not in B = 3 – 2 = 1
So minimum number of people in committee A = 1 + 1 + 2 = 4

### Q4

A girl has to make pizza with different toppings. There are 8 different toppings. In how many ways can she make pizzas with 2 different toppings?
A. 16 B. 56 C. 122 D. 28

D
The girl has to use any two toppings out of 8 available toppings.
Number of ways of selecting r objects from n objects = $^{n}{c}_{r}$
Number of ways to select 2 out of $8 = ^{8}{c}_{2} = \frac {8!}{(2! × 6!)} = 28$

### Q5

A pizza shop made pizzas with many flavors. There are 10 different flavors including Mushroom, in that 7 flavors are taken to make pizza and in which Mushroom topping is necessary. In how many ways they can arrange?
A. 84 B. 120 C. 65 D. 210

A
Given that Mushroom is compulsory and 6 flavors have to be selected from remaining 9 flavors.
Number of ways of selecting r objects from n objects = $^{n}{c}_{r}$
Number of ways to select 6 out of 9 choices = $^{9}{c}_{6} = \frac {9!}{(6! × 3!)} = 84$ ways