Question

Six apples and six mangoes are to be distributed among 10 boys so that each boy receives at least one fruit. Find the number of ways in which the fruits can be distributed.

Answer 1

We solve this problem first by distributing all the 10 students 1 fruit each. Then we can find the number of ways of distributing the remaining 2 fruits among 10 students.
First we take all the possibilities of distributing the 2 fruits among 10 boys so that they get at least one fruit. The possibilities are
(1) Distributing 2 fruits to 2 boys
(2) Distributing 2 fruits to 1 boy.
We use the formula that the number of ways of selecting $'r'$ objects from $'n'$ objects is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

Complete step-by-step answer:
We are given that there are a total of 10 boys and 6 apples and 6 mangoes.
Here we can see that there are a total of 10 boys and 12 fruits.
We are asked to find the number of ways for which each boy gets at least one fruit.
Now, let us distribute 1 fruit to each boy.
Here, we can say that 10 fruits are distributed among 10 boys.
So, we can see that the remaining fruits are 2.
Now, let us take all the possibilities where we can distribute 2 fruits to 0 boys such that each boy gets at least one fruit. The possibilities are
(1) Distributing 2 fruits to 2 boys
(2) Distributing 2 fruits to 1 boy.
Now, let us find the number of ways in each possibility.
(1) Distributing 2 fruits to 2 boys
Let us assume that the number of ways of distributing in this possibility as ${{n}_{1}}$
Here we can see that the number of ways of distributing 2 fruits to 2 boys is nothing but selecting 2 boys from 10 boys because the order is not important.
We know that the formula that the number of ways of selecting $'r'$ objects from $'n'$ objects is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
By using the above formula we get the number of ways in selecting 2 boys is given as
$\Rightarrow {{n}_{1}}={}^{10}{{C}_{2}}$
By using the combinations formula that is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ to above equation we get

& \Rightarrow {{n}_{1}}=\dfrac{10!}{2!\left( 10-2 \right)!} \\\ & \Rightarrow {{n}_{1}}=\dfrac{10\times 9\times 8!}{2\times 8!} \\\ & \Rightarrow {{n}_{1}}=45 \\\ \end{aligned}$$ (2) Distributing 2 fruits to 1 boy. Let us assume that the number of ways of distributing in this possibility as $${{n}_{2}}$$ Here we can see that the number of ways of distributing 2 fruits to 1 boy is nothing but selecting 1 boy from 10 boys because the order is not important. By using the selection formula we get the number of ways in selecting 2 boys is given as $$\Rightarrow {{n}_{2}}={}^{10}{{C}_{1}}$$ By using the combinations formula that is $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$ to above equation we get $$\begin{aligned} & \Rightarrow {{n}_{2}}=\dfrac{10!}{1!\left( 10-1 \right)!} \\\ & \Rightarrow {{n}_{2}}=\dfrac{10\times 9!}{1\times 9!} \\\ & \Rightarrow {{n}_{2}}=10 \\\ \end{aligned}$$ Let us assume that the total number of ways of distributing 12 fruits among 10 boys as $$N$$ then we get $$\Rightarrow N={{n}_{1}}+{{n}_{2}}$$ By substituting the required values in above equation we get $$\begin{aligned} & \Rightarrow N=45+10 \\\ & \Rightarrow N=55 \\\ \end{aligned}$$ Therefore we can conclude that there are a total of 55 ways of distributing 6 apples and 6 mangoes among 10 boys. **Note:** We have a shortcut or solution to this problem. We have the remaining of 2 fruits after distributing 1 fruit to each boy. We have the direct formula that distribution of $$'r'$$ objects among $$'n'$$ people so that each person gets at least 1 object in which each person has already got 1 object is given as $${}^{r+n-1}{{C}_{n-1}}$$ By using the above formula we get the total number of ways of distributing 2 fruits among 10 boys $$\begin{aligned} & \Rightarrow N={}^{10+2-1}{{C}_{10-1}} \\\ & \Rightarrow N={}^{11}{{C}_{9}} \\\ \end{aligned}$$ By using the combinations formula that is $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$ to above equation we get $$\begin{aligned} & \Rightarrow N=\dfrac{11!}{9!\left( 11-9 \right)!} \\\ & \Rightarrow N=\dfrac{11\times 10\times 9!}{2\times 9!} \\\ & \Rightarrow N=55 \\\ \end{aligned}$$ Therefore we can conclude that there are a total of 55 ways of distributing 6 apples and 6 mangoes among 10 boys.