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 is

\(={^8C_3}=56\)

Number of times a child will go to the zoological garden = Number of times he is accompanied by two other

\(=1\times {^7C_2}=21\)

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

$=56−21= 35$

**Edit:** For an alternative approach, check comment by **Sravan Reddy.**

**Sravan Reddy**

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**PRATYUSH ANAND**

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This is really good question. But as it's topic is Algebra never thought of computation.

Alternate way to approach 2nd option in case ${^7C_2}$ does not strike (may take similar or more time):

=> Total number of times he will go to zoo - 56.

=> Total children taken including repetitions - 56*3

=> Due to symmetry, each child will go $\dfrac{56*3}{8} = 21 \text{ times}$