Practice Question on permutation and combination for IIT-JEE , NDA and Airforce

permutation and combination

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Question no. 1

how many words can be formed using all the letters pf the words 'NATION' so that all the three vowels should never come together?

looks_one 354

looks_two 348

looks_3 288

looks_4 None of these

Option looks_3 288

Solution :
No. of ways so that all the three vowels should never come together = Total of ways - No. of ways so that all the three vowels should come together
= 6!/2! - 3! * 3! * 2 = 288

Question no. 2

In how many ways can the letters of the words 'GLOOMY' be arranged so that the two O's should not be together ?

looks_one 240

looks_two 480

looks_3 600

looks_4 720

Option looks_one 240

Solution :
Total number of ways so that the two O's should not be together = total no. of ways - no. of ways two O's should be together
= 6!/2! - 5 * 4! = 240

Question no. 3

If P(77,31) = x and C(77,31 ) = y , then which one of the following is correct ?

looks_one x=y

looks_two 2x = y

looks_3 77x = 31y

looks_4 x > y

Option looks_4 x > y

Solution :

Question no. 4

The number of permutations that can be formed from all the letters of the words 'BASEBALL' is

looks_one 540

looks_two 1260

looks_3 3780

looks_4 5040

Option looks_4 5040

Solution :
Total letters = 8
Repeated letters = 2 , 2 , 2
total no, of ways = 8! / (2! 2! 2!)
= 5040

Question no. 5

What is the number of ways that 4 boys and 3 girls can be seated so that boys and girls alternate ?

looks_one 12

looks_two 72

looks_3 120

looks_4 144

Option looks_4 144

Solution :
If we start with Girl in seating , the arrangement will be G B G B G B B , in this arrangement Two boys are not seated in alternating seat
let's check another arrange : B G B G B G B , This is the right arrangement , Now the ways of ways in which boys and girls can be arranged in this form
No. of ways for each seat from left to right = 4 * 3 * 3 * 2 * 2 * 1 * 1 = 144

Question no. 6

What is the value of ?

looks_one 2ⁿ -1

looks_two 2ⁿ

looks_3 2^n-1

looks_4 2ⁿ + 1

Option looks_one 2ⁿ -1

Solution :

Question no. 7

A,B,C,D and E are coplanar points and three of them lie in a straight line. What is the maximum number of triangles that can be drawn with these points as their vertices?

looks_one 5

looks_two 10

looks_3 9

looks_4 12

Option looks_3 9

Solution :
No. of triangles with n point in which p points are collinear = C(n,3) - C(p,3)
C(5,3) - C(3,3) = 10 - 1 = 9

Question no. 8

Using the digits 1,2,3,4 and 5 only once , how many numbers greater than 41000 can be formed ?

looks_one 41

looks_two 48

looks_3 50

looks_4 55

Option looks_two 48

Solution :
At the first position from left , only two number can be used = 4 , 5
At the second position from left , only four number can be used
At the third position from left , only three number can be used
At the fourth position from left , only two number can be used
At the last position from left , only one number can be used
Total number of ways = 2 * 4 *3 *2 * 1 = 48

Question no. 9

what is the value of n , if P(15, n-1) : P(16,n-2) = 3 : 4

looks_one 10

looks_two 12

looks_3 14

looks_4 15

Option looks_3 14

Solution :

Question no. 10

In how many ways 6 girls can be seated in two chairs?

looks_one 10

looks_two 15

looks_3 24

looks_4 30

Option looks_4 30

Solution :
no. of ways girls can be seated in first chair = 6
no. of ways girls can be seated in second chair = 5
Total number of ways = 5 * 6 = 30

Question no. 11

5 books are to be chosen from a lot of 10 books. If m is the number of ways of choice when one specified book is always included and n is the number of ways of choice when a specified book is always excluded, then which one of the following is correct ?

looks_one m>n

looks_two m=n

looks_3 m = n-1

looks_4 m= n-2

Option looks_two m=n

Solution :
No. of ways when one specified book is always included , we have now choice of 4 books from 9 books
Ways of choice (m) = C(9,4)
No. of ways when one specified book is always excluded , we have now choice of 5 books from 9 books
Ways of choice (n) = C(9,5)
But as we know C(n,r) = C(n , n-r) , therefore C(9,4) = C(9,5) ⇒ m =n

Question no. 12

What is the total number of combination of n different things taken 1,2,3,....n at a time ?

looks_one 2ⁿ⁺¹

looks_two 2²ⁿ⁺¹

looks_3 2^n-1

looks_4 2ⁿ - 1

Option looks_4 2ⁿ - 1

Solution :
Combination if 1 different thing taken at a time = C(n ,1)
Combination if 2 different thing taken at a time = C(n ,2)
Combination if 3 different thing taken at a time = C(n ,3)
.
.
.
Combination if n different thing taken at a time = C(n ,n)
Total number of combination = C(n,1) + C(n,2) + ........ + C(n,n)
2ⁿ = C(n,0) + C(n,1) + C(n,2) + ........ + C(n,n) ⇒ 2ⁿ = 1 + C(n,1) + C(n,2) + ........ + C(n,n)
C(n,1) + C(n,2) + ........ + C(n,n) = 2ⁿ - 1

Question no. 13

In how many ways can a committee consisting of 3 men and 2 women be formed from 7 men and 5 women?

looks_one 45

looks_two 350

looks_3 700

looks_4 4200

Option looks_two 350

Solution :
3 men from 7 men = C(7,3)
2 women from 5 women = C(5,2)
No. of ways = C(7,3).C(5,2) = 7!/(3!4!) . 5!/(2!3!) = 350

Question no. 14

What is the number of signals that can be sent by 6 flags of different colors taking one or more at a time ?

looks_one 21

looks_two 63

looks_3 720

looks_4 1956

Option looks_4 1956

Solution :
Signal using One flag(Six flags) = 6
Signal using two flags = 6 * 5 = 30
Signal using three flags = 6 * 5 * 4 = 120
Signal using four flags = 6 * 5 * 4 * 3 = 360
Signal using five flags = 6 * 5 * 4 * 3 *2 = 720
Signal using Six flags = 6 * 5 * 4 * 3 * 2 * 1 = 720
total number of ways = 6 + 30 + 120 + 360 + 720 + 720 = 1956

Question no. 15

What is the number of words that can be formed from the letters of the words 'UNIVERSAL' the vowels remaining always together ?

looks_one 720

looks_two 1440

looks_3 17280

looks_4 21540

Option looks_3 17280

Solution :
No. of vowels = 4
Ways for Four vowels can be arranged = 4!
Ways of remaining letters = 5!
there are 6 positions in which 4 vowels can be placed in the Universal letters
Total number of ways = 4! * 6! * 6 = 17280

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