Aptitude Discussion

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. |

**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.