Permutation and Combination Questions for MAH-CET

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Permutation & Combination Questions
Permutation & Combination Questions

Permutation and Combination Questions for MAH-CET

Here you can download a free Permutation and Combination questions PDF with answers for MAH MBA CET by Cracku. These are some tricky questions in the MAH MBA CET exam that you need to find Permutation and Combination answers for the given questions. These questions will help you to make practice and solve the Permutation and Combination questions in the MAH MBA CET exam. Utilize this best PDF practice set which is included answers in detail. Click on the below link to download the Permutation and Combination MCQ PDF for MBA-CET 2022 for free.

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Question 1: Letters of the word DIRECTOR are arranged in such a way that all the vowel come together .Find the No of ways making such arrangement?

a) 4320

b) 720

c) 2160

d) 120

e) None of these

1) Answer (A)

Solution:

Word – DIRECTOR

So “I,E,O” are there are 3! ways to arrange the vowels

Now “D,R,C,T,R” are the remaining alphabets ,

Condition is that the vowels should always be together so we can assume the vowels as a single alphabet/unit say “X” (‘X’=’I,E,O’) so now we have a new word – “D,R,C,T,R,X”

Possible arrangements for this word = 6!

Thus total number of ways to rearrange DIRECTOR with vowels grouped together = (Possible arrangements of ‘DRCTRX’) $\times$ (Possible arrangements of vowels)

= 6! $\times$ 3! = $720 \times 6 = 4320$

=> Ans – (A)

Question 2: In how many different ways can the letters of the word TRUST’ be arranged?

a) 240

b) 120

c) 80

d) 25

e) None of these

2) Answer (E)

Solution:

Word = ‘TRUST’

There are 5 letters and ‘T’ is repeated.

If there are ‘n’ letters and ‘r’ are repeating, then number of ways of arranging them = $\frac{n !}{r !}$

$\therefore$ Number of ways in which letters of the word ‘TRUST’ can be arranged

= $\frac{5 !}{2 !} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}$

= $5 \times 4 \times 3 = 60$

=> Ans – (E)

Question 3: How many such pairs of letters are there in the word FOREHAND each of which have as many letters between them in the word as they have in the English alphabet?

a) None

b) One

c) Two

d) Three

e) More than three

3) Answer (C)

Solution:

Word = FOREHAND

There are 2 pairs of letters which have as many letters between them in the word as they have in the English alphabet

FA and RN

Question 4: If it is possible to make only one meaningful word from the second, the fourth, the sixth and the ninth letters of the word PROACTIVE, using each letter only once, second letter of that word is your answer. If more than one word can be formed your answer is M and if no such word can be formed your answer is N.

a) A

b) E

c) T

d) M

e) N

4) Answer (D)

Solution:

Word = PROACTIVE

2nd, 4th, 6th and 9th letters = R, A, T, E

No. of words that can be formed by using (R,A,T,E)

= Rate , Tear

Since only 2 words are formed

=> Ans = M

Question 5: In how many different ways can the letters of the word DRASTIC be arranged in such a way that the vowels always come together ?

a) 720

b) 360

c) 1440

d) 540

e) None of these

5) Answer (C)

Solution:

There are 7 letters in the word ‘DRASTIC’ including 2 vowels (A,I) and 5 consonants (D,R,S,T,C).

Considering the two vowels as 1 letter, we have 6 letters which can be arranged in 6! ways

But  corresponding to each way of these arrangements, the vowels can be put together in 2! ways.

Hence, required number of words = $6! \times 2!$

= 720 * 2 = 1440

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Question 6: Which of the following set of fractions is in ascending order ?

a) 13/15, 11/13, 7/8, 8/9

b) 11/13, 13/15, 7/8, 8/9

c) 8/9, 7/8, 13/15, 11/13

d) 7/8, 8/9, 11/13, 13/15

e) None of these

6) Answer (B)

Solution:

Decimal equivalent of :

$\frac{13}{15} = 0.87$

$\frac{11}{13} = 0.85$

$\frac{7}{8} = 0.875$

$\frac{8}{9} = 0.89$

Clearly, 0.85 < 0.87 < 0.875 < 0.89

=> $\frac{11}{13} < \frac{13}{15} < \frac{7}{8} < \frac{8}{9}$

Question 7: Certain number of pieces of an article are to be packed in boxes, such that each box contains 145 pieces. If after packing them in 32 boxes 25 pieces are left out, what was the number of pieces to be packed ?

a) 4566

b) 4655

c) 4465

d) 4640

e) None of these

7) Answer (E)

Solution:

No. of pieces

= 32 * 145 + 25

= 4640 + 25 = 4665

Question 8: In how many different ways can the letters of the word ‘HAPPY’ be arranged ?

a) 120

b) 140

c) 60

d) 70

e) None of these

8) Answer (C)

Solution:

The word HAPPY contains 5 letters in which letter ‘P’ comes twice.

=> Number of arrangements = $\frac{5!}{2!}$

= $\frac{120}{2} = 60$

Question 9: In how many different ways can the letters of the word ‘FRANCE’ be arranged ?

a) 2400

b) 720

c) 2005

d) 5040

e) None of these

9) Answer (B)

Solution:

The word FRANCE consists of 6 distinct letters

=> Required number of arrangements = 6!

= 720

Question 10: In how many different ways can the letters of the word ‘CREAM’ be arranged?

a) 720

b) 240

c) 360

d) 504

e) None of these

10) Answer (E)

Solution:

The word CREAM consists of five distinct letters

=> No. of ways the letters will be arranged = $5!$

= 120

Question 11: In how many different ways can the letters of the word “ARISE’ be arranged?

a) 90

b) 60

c) 180

d) 120

e) None of these

11) Answer (D)

Solution:

The word ‘ARISE’ has 5 different letters

=> Required number of arrangements = $5!$

= $5 \times 4 \times 3 \times 2 \times 1 = 120$

Question 12: In how many different ways can the letters of the world ‘SECOND’ be arranged?

a) 720

b) 120

c) 5040

d) 270

e) None of these

12) Answer (A)

Solution:

The word ‘SECOND’ has 6 different letters

=> Required number of arrangements = $6!$

= $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

Question 13: In how many different ways can the letters of the word “PRIDE” be arranged ?

a) 60

b) 120

c) 15

d) 360

e) None of these

13) Answer (B)

Solution:

The word ‘PRIDE’ consists of 5 distinct letters

=> Number of arrangements = $5!$

= $5 \times 4 \times 3 \times 2 \times 1 = 120$

Question 14: In how many different ways can the letters of the word ‘REPLACE’ be arranged ?

a) 2630

b) 5040

c) 1680

d) 2580

e) None of these

14) Answer (E)

Solution:

The word ‘REPLACE’ has 2 E,1 A,1 R,1 C,1 L,1 P.
Hence no. of ways in which it can be arranged=$\frac{7!}{2!}$.
=$\frac{5040}{2}$.
=$2520$.
Hence, Option E is correct

Question 15: In how many different ways can the letters of the word ‘QUOTED’ be arranged

a) 720

b) 360

c) 1440

d) 320

e) None of these

15) Answer (A)

Solution:

Total number of arrangement= 6!=720

Question 16: In how many different ways can the letters of word ‘POWERS’ he arranged ?

a) 640

b) 720

c) 1440

d) 360

e) None of these

16) Answer (B)

Solution:

There are 6 different letters in the word POWERS .
The can be arranged in 6! ways = 720

Question 17: What would be the area of a circle whose diameter is 35 cms ‘?

a) 962.5 sq.cm.

b) 875.5 sq.cm.

c) 981.5 sq. cm .

d) 886.5 sq.cm.

e) None of these

17) Answer (A)

Solution:

We know that, $Area = \frac{\pi d^{2}}{4}$
$ = \frac{3.14 * 35^{2}}{4}$

$=962.5 sq.cm$

Option A is the correct answer.

Question 18: In how many different ways can the letters of word ‘REMAKE’ be arranged ?

a) 720

b) 130

c) 360

d) 180

e) None of these

18) Answer (C)

Solution:

The total number of alphabets in the word REMAKE is 6.

So the word can be rearranged in 6! ways.

But the alphabet “E” appears twice in the word.

Hence the word can be rearranged only in $\frac{6!}{2!}$ ways.

$\frac{6!}{2!}$=$\frac{720}{2}$

=360.

Hence Option C is the correct answer.

Question 19: In how many different ways can the letters of the word ‘MARKERS’ be arranged ?

a) 840

b) 5040

c) 2520

d) 1680

e) None of these

19) Answer (C)

Solution:

The word MARKERS has 7 letters in which ‘R’ comes twice.

∴ Required number of arrangements = $\frac{7!}{2!}$

= 7*6*5*4*3 = 2520

Question 20: A committee of 6 members is to be selected from a group of 8 men and 6 women in such as way that at least 3 men are there in the committee. In how many different ways can it be done ?

a) 2506

b) 2534

c) 1120

d) 1050

e) None of these

20) Answer (B)

Solution:

There are a total of 8 men and 6 women.

Number of ways in which the committee has exactly 3 men is $^8C_3 \times ^6C_3 = 1120$
Number of ways in which the committee has exactly 4 men is $^8C_4 \times ^6C_2 = 1050$
Number of ways in which the committee has exactly 5 men is $^8C_5 \times ^6C_1 = 336$
Number of ways in which the committee has exactly 6 men is $^8C_6 \times ^6C_0 = 28$

Hence, the total is $1120 + 1050 + 336 + 28 = 2534$

Question 21: Two girls and four boys are to be seated in a row, in such a way that the girls do not sit together. In how many different ways can it be done ?

a) 720

b) 480

c) 360

d) 240

e) None of these

21) Answer (B)

Solution:

In order to solve this question, let us find the total number of ways of arranging four boys and two girls and remove the number of arrangements in which both the girls sit together.

Total number of ways of arranging 4 boys and 2 girls is 6! = 720
Total number of ways of arranging them in such a way that both the girls sit together is (6-1)! * 2 = 240

Hence, the answer is 720 – 240 = 480

Question 22: In how many different ways can letters of the word “BELIEVE” be arranged ?

a) 840

b) 1680

c) 2520

d) 5040

e) None of these

22) Answer (A)

Solution:

BELIEVE
The total number of alphabets in the word ‘BELIEVE’ is 7.
There are three ‘E’s in these seven letters.

Hence, the number of different ways the letters of the word ‘BELIEVE’ can be arranged is $\frac{7!}{3!} = 7*6*5*4 = 840$

Question 23: 4 boys and three girls are to be seated in a row in such a way that no two boys sit adjacent to each other. In how many different ways can it be done?

a) 5040

b) 30

c) 144

d) 72

e) None of these

23) Answer (C)

Solution:

3 girls can be seated in 3! ways
The required arrangement is B G B G B G B
4 boys can be seated in 4! ways
Number of required ways = $3! \times 4!$ = 144

Question 24: A committee of 3 members is to be formed out of 3 men and 4 women. In how many different ways can it be done so that at least one member is a woman?

a) 34

b) 12

c) 30

d) 36

e) None of these

24) Answer (A)

Solution:

Number of ways of selecting any 3 members = $^7C_3 = 35$
Number of ways of selecting only men = $^3C_3 = 1$
Number of ways of selecting such that at least one member is a woman = 35 – 1 = 34

Question 25: In how many different ways a group of 4 men and 4 women be formed out of 7 men and 8 women ?

a) 2450

b) 105

c) 1170

d) Cannot be determined

e) None of these

25) Answer (A)

Solution:

The number of ways of selecting 4 men from 7 men is $^7C_4 = 35$
The number of ways of selecting 4 women from 8 women is $^8C_4 = 70$

Hence, the total number of ways is $70 \times 35 = 2450$

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