Permutation and Combination
Question no. 16
What is the number of three digit odd numbers formed by using the digits 1,2,3,4,5,6 if repetition of digits is allowed ?
looks_one 60
looks_two 108
looks_3 120
looks_4 216
Option looks_two 108
Solution :
Number of digits at one's place = 1 , 3 , 5 [Total number is 3]
Number of digits at ten's place = 6 [1,2,3,4,5,6]
Number of digits at hundred place =6 [ 1, 2 , 3 , 4 , 5 , 6 ]
Total number of ways = 3 * 6 * 6 = 108
Number of digits at one's place = 1 , 3 , 5 [Total number is 3]
Number of digits at ten's place = 6 [1,2,3,4,5,6]
Number of digits at hundred place =6 [ 1, 2 , 3 , 4 , 5 , 6 ]
Total number of ways = 3 * 6 * 6 = 108
Question no. 17
What is the number of ways of arranging the letters of the word "BANANA" so that no two N's appear together?
looks_one 60
looks_two 40
looks_3 80
looks_4 100
Option looks_two 40
Solution :
Total no. of ways = 6!/(3!2!) = 60
Two N's come together = 4!*5 / 3! = 20
No. of ways no two N's appear together = 60 - 20 = 40
Total no. of ways = 6!/(3!2!) = 60
Two N's come together = 4!*5 / 3! = 20
No. of ways no two N's appear together = 60 - 20 = 40
Question no. 18
How many times does the digit 3 appear while writing the integers from 1 to 1000?
looks_one 269
looks_two 308
looks_3 300
looks_4 None of these
Option looks_4 None of these
Solution :
3 occur at exactly 3 times (333) = 1
3 occur at exactly 2 times = 3C2 * 9 = 27
3 occur at exactly 1 time = 3C1 * 9 * 9 = 243
Total no. of ways = 1 + 27 + 243 = 271
3 occur at exactly 3 times (333) = 1
3 occur at exactly 2 times = 3C2 * 9 = 27
3 occur at exactly 1 time = 3C1 * 9 * 9 = 243
Total no. of ways = 1 + 27 + 243 = 271
Question no. 19
If C(n ,12) = C(n ,8) , then whats is the value of C( 22, n) ?
looks_one 131
looks_two 231
looks_3 256
looks_4 292
Option looks_two 231
Solution :


Question no. 20
What is the number of words formed from the letters of the word 'JOKE' so that the vowels and consonants alternate ?
looks_one 4
looks_two 8
looks_3 12
looks_4 None of these
Option looks_two 8
Solution :
arrangement of vowel and consonant at alternate = V C V C , C V C V
For V C V C arrangement
No. of ways = 2 * 2 * 1 * 1 = 4
For C V C V arrangement
No. of ways = 2 * 2*1 *1 = 4
total no. of ways = 8
arrangement of vowel and consonant at alternate = V C V C , C V C V
For V C V C arrangement
No. of ways = 2 * 2 * 1 * 1 = 4
For C V C V arrangement
No. of ways = 2 * 2*1 *1 = 4
total no. of ways = 8
Question no. 21
What is the value of r , if P(5,r) = P(6,r-1) ?
looks_one 9
looks_two 4
looks_3 5
looks_4 2
Option looks_two 4
Solution :


Question no. 22
What is the smallest natural number n such that n! is divisible by 990?
looks_one 9
looks_two 11
looks_3 33
looks_4 99
Option looks_two 11
Solution :
990 is the factor of n!
990 has factor 11 * 10 * 9
So, n! must contain 11 , 10 , 9
only 11 is smallest number which contain all the number
990 is the factor of n!
990 has factor 11 * 10 * 9
So, n! must contain 11 , 10 , 9
only 11 is smallest number which contain all the number
Question no. 23
If P(32,6)= kC(32,6), then what is the value of k?
looks_one 6
looks_two 32
looks_3 120
looks_4 720
Option looks_4 720
Solution :


Question no. 24
From 7 men and 4 women a committee of 6 is to be formed such that the committee contains at least two women . what is the number of ways to do this?
looks_one 210
looks_two 371
looks_3 462
looks_4 5544
Option looks_two 371
Solution :
At least 2 women
No. of ways = 7C4 * 4C2 + 7C3 * 4C3 + 7C2 * 4C4
35* 6 + 35*4 + 21 = 210 + 140 + 21 = 371
At least 2 women
No. of ways = 7C4 * 4C2 + 7C3 * 4C3 + 7C2 * 4C4
35* 6 + 35*4 + 21 = 210 + 140 + 21 = 371
Question no. 25
How many words , with or without meaning by using all the letters of the word 'MACHINE' so that the vowels occurs only the odd positions?
looks_one 1440
looks_two 720
looks_3 640
looks_4 576
Option looks_4 576
Solution :
No. of vowels = 3 and No. of letter = 7
4 places and 3 vowels = 4P3 = 4 !
remaining letters can be arranges in ways = 4 !
Total no. of ways = 4! * 4 ! = 576
No. of vowels = 3 and No. of letter = 7
4 places and 3 vowels = 4P3 = 4 !
remaining letters can be arranges in ways = 4 !
Total no. of ways = 4! * 4 ! = 576
Question no. 26
In how many ways can 3 books on Hindi and 3 books on English be arranged in a row on a shelf , so that not all the Hindi books are together?
looks_one 144
looks_two 360
looks_3 576
looks_4 720
Option looks_3 576
Solution :
Total no. of ways book can be arranged = 6! = 720
No. of ways all the hindi books are together = 3! * 4! = 144
No. of ways not all the Hindi books are together = 720-144 = 576
Total no. of ways book can be arranged = 6! = 720
No. of ways all the hindi books are together = 3! * 4! = 144
No. of ways not all the Hindi books are together = 720-144 = 576
Question no. 27
If 7 points out of 12 are in the same straight line, then what is the number of triangles formed?
looks_one 84
looks_two 175
looks_3 185
looks_4 201
Option looks_3 185
Solution :
No. of triangles = nC3 -pC3 where n is total no. of points and p is the collinear points
No. of triangles = 12C3 -7C3
No. of triangles =
No. of triangles = nC3 -pC3 where n is total no. of points and p is the collinear points
No. of triangles = 12C3 -7C3
No. of triangles =
Question no. 28
A group consists of 5 men and 5 women. If the number of 27 different five person committees containing k men and 5-k women is 100 , what is the value of k ?
looks_one 2 only
looks_two 3 only
looks_3 2 or 3
looks_4 4
Option looks_3 2 or 3
Solution :


Question no. 29
What is the number of five- digit numbers formed with 0,1,2,3,4 without any repetition of digits?
looks_one 24
looks_two 48
looks_3 96
looks_4 120
Option looks_3 96
Solution :
No. of 5th place digit = 4
No. of 4th place digit = 4
No. of 3rd place digit = 3
no. of 2nd place digit = 2
No. of first place digit = 1
Total no. of ways = 4 * 4 * 3 * 2 * 1 = 96
No. of 5th place digit = 4
No. of 4th place digit = 4
No. of 3rd place digit = 3
no. of 2nd place digit = 2
No. of first place digit = 1
Total no. of ways = 4 * 4 * 3 * 2 * 1 = 96
Question no. 30
The number of ways in which 5 beads of different colours form a necklace is
looks_one 12
looks_two 24
looks_3 120
looks_4 60
Option looks_3 120
Solution :
No. of ways = 5* 4 * 3 * 2 * 1 = 120
No. of ways = 5* 4 * 3 * 2 * 1 = 120
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