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Download CMAT 2022 Permutations & Combinations Questions pdf by Cracku. Very Important Permutation and Combination Questions for CMAT 2022 based on asked questions in previous exam papers. These questions will help your CMAT preparation. So kindly download the PDF for reference and do more practice.

Question 1: In how many ways can the letters of the word “EXAMPLE” be arranged such that all the vowels are together?

a) 480

b) 120

c) 360

d) 240

e) None of the above

Question 2: In how many ways can the letters of the word “MASTER” be arranged?

a) 720

b) 360

c) 480

d) 120

e) None of the above

Question 3: A man has 8 friends – 4 men and 4 women. He has to invite at least 6 of them to a party such that exactly 3 women attend. In how many ways can he invite the friends?

a) 48 ways

b) 25 ways

c) 20 ways

d) 36 ways

e) None of the above

Question 4: How many different words can be made with the letters R,H,T,Y,I,P,Q,L?

a) 40320

b) 5040

c) 720

d) Can’t be determined

e) None of the above

Question 5: There is a class of 50 of which 35 are girls. In how different ways can a group of 10 girls and 15 boys be picked for a sports event?

a) $^{35}C_{10}$

b) $^{35}P_{10}$

c) $^{35}C_{10}$*$^{50}C_{15}$

d) Can’t be determined

e) None of the above

Question 6: In how many different ways can the letters of the word PARENTS be arranged?

a) 2000

b) 4000

c) 8000

d) 16000

e) None of the above

Question 7: Which of the following words can be arranged in 720 different ways?

a) NECKLACE

b) SCAR

c) JUSTIN

d) SCRIB

e) TURNOFF

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

a) 10080

b) 20160

c) 40320

d) 80720

e) 161440

Question 9: In how many different ways can the letters of the word STRETCH be rearranged using all the letters each time?

a) 2000

b) 3000

c) 5400

d) 6000

e) None of the above

Question 10: Find the no of parallelograms that can be formed from a set of four parallel lines intersecting from a set of four parallel lines intersecting another set of three parallel lines?

a) 12

b) 24

c) 48

d) 18

e) None of these

Question 11: In how many different ways a group of 4 men and 4 women can be formed out of 7 men and 8 women?

a) 2450

b) 1050

c) 117

d) 232

e) None of these

Question 12: Find the number of different signals that can be transmitted by arranging 3 yellow flags, 4 red flags and 2 blue flags on a pole. All the flags are used to transmit the signal.

a) 1200

b) 1260

c) 1300

d) 1350

e) None of these

Question 13: The number of arrangements that can be made with the letters of the word ‘MATHS’. So, that Letter M will occupy always the first place is?

a) 3!

b) 4!

c) 5!

d) 6!

e) None of these

Question 14: Find the number of different words that can be formed from the word ‘SUCCEED’

a) 1390

b) 1400

c) 1320

d) 1260

e) None of these

Question 15: There are 10 students in a batch. In how many ways can the first five ranks be getting in?

a) 30000

b) 30240

c) 30500

d) 30600

e) None of these

Question 16: In how many ways we can arrange the letters of the word “VICTORY” so that no two vowels come together?

a) 4500

b) 3600

c) 4000

d) 4400

e) 4800

Question 17: How many words of five letters can be formed from the word PRESS without any repetition of alphabets?

a) 120

b) 60

c) 30

d) 40

e) 90

Question 18: What is the number of ways in which we can select 3 students in a class of 40 students?

a) 8990

b) 9880

c) 5999

d) 3999

e) 5040

Question 19: In how many ways can Alex and his 5 friends be seated around a circular table if clockwise and anti-clockwise directions are considered different?

a) 120

b) 720

c) 60

d) 360

e) None of the above

Question 20: There are 4 letters:A,B,C,D. In how many ways can we arrange the four letters such that B always comes before C.

a) 20

b) 8

c) 12

d) 16

e) None of these

The vowels are E, E, A. The consonants are X, M, P, L. Consider all the vowels to be one unit. The number of ways of arrangement of the 5 units (4 consonants + 1 vowels unit) = 5! ways = 120 ways. The vowels can be arranged within themselves in 3!/2! ways = 3 ways. So, the total number of ways of arrangement = 120 * 3 = 360 ways

The total number of letters is 6 and no letter is repeated. So, the total number of arrangements is 6! = 720 ways.

He can invite 3 women and 3 men or 3 women and 4 men. 3 women can be invited in $^4C_3$ ways = 4 ways. 3 men can be invited in 4 ways and 4 men can be invited in 1 way. Total number of ways = 4*(4+1) = 20 ways

There are 8 letters that are distinct. So, the number of combinations = 8!

All 15 boys have to be selected. So, they don’t add to the combinations. Number of ways of selecting girls is calculated using $^{n}C_r$

The word “Parents” has 7 letters. So, the number of ways to arrange it are 7! = 5040 ways.

720 is 6! Since JUSTIN has 6 letters, it can be arranged in 720 ways.

The word “Question” has 8 letters. So, the number of ways to arrange it are 8! = 40,320 ways.

The letter T repeats twice. So, the number of unique words that can be formed = 7!/2!

Required number = $^4 C_2 * ^3 C_2$

$^4 C_2 = \frac{4!}{2!(4-2)!}=\frac{4\times3\times2\times1}{2!\times2!} =6$

$^3 C_2 = \frac{3!}{1!2!} =3$

So, $^4 C_2 * ^3 C_2 = 6\times3 = 18$

4 men out of 7 men and 4 women out of 8 women can be formed in
$^7C_4 * ^8C_4$ ways
That is, (7×6×5×4)/(1×2×3×4)×((8×7×6×5)/(1×2×3×4))=35×70=2450 ways

Total number of flags = 9
Number of different signals which can be transmitted = 9!/(3! 4! 2!) = (9 ×8 ×7 ×6 ×5 ×4!)/(3 ×2 ×1 ×2 ×1 ×4! ) = 9 × 4 × 7 × 5 = 1260

M A T H S
When we fix ‘M’ at the first palce, the word can be of the form M _ _ _ _
Now, we have 4 alternatives for four blanks. These can be permuted as 4!

=> Number of ways = 4!

Number of letters = 7 = n
Number of C = 2 = p
Number of E = 2 = q
n!/(p! q!) = 7!/2!2! = (7 ×6 ×5 ×4 ×3 ×2!)/(2! ×2 ×1) = 7 × 6 × 5 × 2 × 3 = 1260

Number of students = 10
First 10 ranks have to be get = r = 5
$^nP_r$ = n!/(n -r)!
$^{10}P_5$ = 10!/(10 -5)! = (10 ×9 ×8 ×7 ×6 ×5!)/5! = 10 × 9 × 8 × 7 × 6 = 30240
Value of $^{10}P_5$ = 30240

Total no. of ways arranging all 7 letters of the word = 7!
Total no. of ways arranging letters with combining both vowels together = 6!*2
Ways of arranging letters when no two vowels come together = 7! – 6!*2 = 3600

The number of alphabets in PRESS is 5 of which there are two alphabets which are the same (S). Hence, the total number of permutations possible is $\frac{5!}{2!}=\frac{5*4*3*2*1}{2*1}=60$

40C3 = 40!/37!*3! = 9880

Number of ways =$^4C_2*2!$ = 12.