Question

Fruits are to be distributed amongst $11$ children from a basket of fruits containing $5$ mangoes, $4$ apples and $2$ bananas. Each child is to get one fruit. In how many ways the fruits can be distributed?

Answer 1

Hint: The number of ways of choosing $r$ things from $n$ things is given by ${}^{n}{{C}_{r}}$.
The value of ${}^{n}{{C}_{r}}$ is given as ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$.

Given- Total number of children $=11$.
Total number of mangoes $=5$.
Total number of bananas $=2$.
Total number of apples $=4$.
We need to find the number of ways to distribute the fruits among the children such that each child gets one fruit.
Now , instead of choosing fruits to divide among the children , we will choose children to whom each fruit is distributed . This is done because each child is different and there will be $11$ ways to choose each child independently , but each fruit is similar in a group , i.e. each mango is similar to another , each apple is similar to another . So , there will be one way to choose one mango from \[5\] mangoes.
 So , we need to choose $5$ children for mangoes, $4$ children for apples and $2$ children for bananas.
We know, number of ways of choosing $r$ things from $n$ things is given by ${}^{n}{{C}_{r}}$.
And we also know ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
So , number of ways of choosing $5$ children for mangoes from $11$ children $={}^{11}{{C}_{5}}$
So , remaining children $=11-5=6$
Now, $4$ children are to be chosen from $6$ children for apples. So , number of ways $={}^{6}{{C}_{4}}$
Remaining number of children $=6-4=2$
Number of ways of choosing $2$ children from $2$ children for bananas $={}^{2}{{C}_{2}}$
Now, all the selections are dependent on each other. So , total number of ways of choosing
\[\begin{align}
  & ={}^{11}{{C}_{5}}\times {}^{6}{{C}_{4}}\times {}^{2}{{C}_{2}} \\
 & =\dfrac{11!}{5!\left( 11-5 \right)!}\times \dfrac{6!}{4!\left( 6-4 \right)!}\times \dfrac{2!}{\left( 2-2 \right)!2!} \\
 & =\dfrac{11!}{5!{6!}}\times \dfrac{{6!}}{{2!}4!}\times \dfrac{{2!}}{0!2!} \\
 & =\dfrac{11!}{5!}\times \dfrac{1}{4!}\times \dfrac{1}{0!2!} \\
\end{align}\]
 Now, we know $\dfrac{a!}{b!}=\left( a \right)\times \left( a-1 \right)\times \left( a-2 \right)\times .......\times \left( a-b \right)$
And $0!=1$
So,
$\begin{align}
  & =\dfrac{11!}{5!}\times \dfrac{1}{4!}\times \dfrac{1}{0!2!}=11\times 10\times 9\times 8\times 7\times 6\times \dfrac{1}{4\times 3\times 2\times 1}\times \dfrac{1}{1\times 2\times 1} \\
 & =11\times 10\times 9\times 7 \\
 & =77\times 90 \\
 & =6930 \\
\end{align}$
Hence , the number of ways of distributing $5$ mangoes, $4$ apples and $2$ bananas among $11$ children is equal to \[6930\].

Note: Alternative Solution:
We know , the number of ways of distributing \[n\] things such that \[p\] are of one kind , \[q\] are of second kind and \[r\] are of third kind is given by \[\dfrac{n!}{p!\times q!\times r!}\]. So , the number of ways of distributing $5$ mangoes, $4$ apples and $2$ bananas among $11$ children is given by \[\dfrac{11!}{5!\times 4!\times 2!}=6930\] .
 Hence , the number of ways of distributing $5$ mangoes, $4$ apples and $2$ bananas among $11$ children is equal to \[6930\].