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A father with 8 children takes 3 at a time to a zoological garden. If the father does not take the same 3 children more than once, then the number of times he will go to the garden.

Last updated date: 22nd Feb 2024
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IVSAT 2024
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Hint: Here, we will consider the concept of combinations, since combination is the selection of all or sets of an objects from a group of objects without considering the order of the objects.

Number of children to the father are 8 and out of them 3 children are taken at a time to the zoological garden.
As we know that the selection of r objects from a set of n objects without any order can be written as ${}^n{C_r}$.Here, the value of r is 3 and the value of n is 8.
$ \Rightarrow r = 3$ and $n = 8$
Now, the number of times the father will go to the garden can be written as ${}^8{C_3}$. Since, he is taking 3 children out of 8 children to the zoological garden.
   \Rightarrow {}^8{C_3} = \dfrac{{8!}}{{3!(8 - 3)!}}[\because {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}] \\
   \Rightarrow {}^8{C_3} = \dfrac{{8!}}{{3!*5!}} = \dfrac{{8*7*6*5!}}{{6*5!}} = 56[\because 8! = 8*7*6*5!] \\ $
Therefore, the number of times the father goes to the garden is 56.
Hence, the correct option is ‘C’.

Note: Since, the question was to find the number of times father goes to the garden we just consider the selection of 3 children out of 8 children as father goes with the children every time.