Question

Out of $5$ apples, $10$ mangoes and $15$ oranges, any $15$ fruits are to be distributed among two persons. Then the total number of ways of distribution?

Answer 1

Hint: Here, in the given question, we need to find the total number of ways of distribution of $5$ apples, $10$ mangoes and $15$ oranges, any $15$ fruits among two persons. To solve the given question, we will use the multinomial theorem. As here we need to find the ways of distributing $15$ fruits, it means we need to find the coefficient of ${x^{15}}$.

Complete step by step answer:
Given, there are $5$ apples, $10$ mangoes and $15$ oranges.
Total number of fruits = $30$
Any $15$ fruits are to be distributed among two persons.
Therefore, number of ways of distributing $15$ fruits = coefficient of ${x^{15}}$ in $\left( {1 + {x^1} + {x^2} + ..... + {x^5}} \right)\left( {1 + {x^1} + {x^2} + ..... + {x^{10}}} \right)\left( {1 + {x^1} + {x^2} + ..... + {x^{15}}} \right)$.
\[ \Rightarrow \left( {\dfrac{{1 - {x^6}}}{{1 - x}}} \right)\left( {\dfrac{{1 - {x^{11}}}}{{1 - x}}} \right)\left( {\dfrac{{1 - {x^{16}}}}{{1 - x}}} \right)\]
This can also be written as:
\[ \Rightarrow \left( {1 - {x^6}} \right)\left( {1 - {x^{11}}} \right)\left( {1 - {x^{16}}} \right){\left( {1 - x} \right)^{ - 3}}\]
On multiplication of terms written first three brackets, we get
\[ \Rightarrow \left( {1 - {x^6} - {x^{11}} + {x^{17}} - {x^{16}} + {x^{22}} + {x^{27}} - {x^{33}}} \right){\left( {1 - x} \right)^{ - 3}}\]
As here we need to find the coefficient of ${x^{15}}$, so let us consider the terms with power of $x$ less than $15$ or equal to $15$ and solve it further.
\[ \Rightarrow \left( {1 - {x^6} - {x^{11}}} \right){\left( {1 - x} \right)^{ - 3}}\]
Let us expand ${\left( {1 - x} \right)^{ - 3}}$ using general term formula $\dfrac{{\left( {r + 1} \right)\left( {r + 2} \right)}}{2}{x^r}$.
\[ \Rightarrow \left( {1 - {x^6} - {x^{11}}} \right)\left( {1 + 3x + 6{x^2} + 10{x^3} + 15{x^4} + .......... + 55{x^9} + .... + 135{x^{15}} + ...} \right)\]
For a coefficient of ${x^{15}}$, we will consider only $135{x^{15}}$, $55{x^9}$ and $15{x^4}$.
\[ \Rightarrow \left( {1 - {x^6} - {x^{11}}} \right)\left( {15{x^4} + 55{x^9} + 135{x^{15}} + ...} \right)\]
For coefficient of ${x^{15}}$, we have
$ \Rightarrow 136 - 55 - 15$
On subtraction of terms, we get
$ \Rightarrow 66$
Therefore, the total number of ways of distribution are $66$.

Note:
Remember that as here we have written $\left( {1 + {x^1} + {x^2} + ..... + {x^5}} \right)$ for apples, here $1$ is nothing but ${x^0}$ (means $0$ apple). While solving this type of questions, always remember that when we multiply two numbers or variables with the same base, we simply add the exponents. We should take care of the calculations so as to be sure of our final answer.