Permutation-Combination
Aptitude

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Q.

A father with $8$ children takes $3$ children at a time to the zoological garden, as often as he can without taking the same $3$ children together more than once. Then:

 A.

number of times he will go to zoological garden is $56$.

 B.

number of times each child will go to the zoological garden is $21$.

 C.

number of times a particular child will not go to the zoological garden is $35$.

 D.

All of the above.

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Solution:
Option(D) is correct

The number of times the father would go to the zoological garden 

$=$ Number of ways of selection of $3$ children taken at a time 

$= ^8C_3=56$

Number of times a child will go to the zoological garden 

$=$ Number of times he is accompanied by two other 

$=1\times ^7C_2=21$

$ \Rightarrow $ Number of times a child will not go to the zoological garden: 

$=56-21= 35$

Thus, all the given option options are correct.


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