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?

• A. $66$
• B. $36$
• C. $60$
• D. None of the above

Answer 1

The correct option is A

$66$

Explanation for correct option:

Calculating the probability of distributing $15$ fruits among two people:

Given, there are $5$ apples, $10$ mangoes and $15$ oranges.

Total number of fruits,

$$\begin{gathered} =5+10+15 \\ =30 \end{gathered}$$

Any $15$ fruits have to be distributed among two persons.

Therefore, the number of ways of distribution = Coefficient of $x^{15}$ in

$$\begin{gathered} \left(1+x+x^2+....+x^5\right)\times \left(1+x+x^2+....+x^{10}\right)\times \left(1+x+x^2+....+x^{15}\right) \\ =\left(1\times 1\times x^{15}\right)+\left(1\times x\times x^{14}\right)+\left(1\times x^2\times x^{13}\right)+\dots \dots \dots \dots +\left(1\times x^{10}\times x^5\right) \\ +\left(x\times 1\times x^{14}\right)+\left(x\times x\times x^{13}\right)+\left(x\times x^2\times x^{12}\right)+\dots \dots \dots \dots +\left(x\times x^{10}\times x^4\right) \\ +\left(x^2\times 1\times x^{13}\right)+\left(x^2\times x\times x^{12}\right)+\left(x^2\times x^2\times x^{11}\right)+\dots \dots \dots \dots +\left(x^2\times x^{10}\times x^3\right) \\ +\left(x^3\times 1\times x^{12}\right)+\left(x^3\times x\times x^{11}\right)+\left(x^3\times x^2\times x^{10}\right)+\dots \dots \dots \dots +\left(x^3\times x^{10}\times x^2\right) \\ +\left(x^4\times 1\times x^{11}\right)+\left(x^4\times x\times x^{10}\right)+\left(x^4\times x^2\times x^9\right)+\dots \dots \dots \dots +\left(x^4\times x^{10}\times x^1\right) \\ +\left(x^5\times 1\times x^{10}\right)+\left(x^5\times x\times x^9\right)+\left(x^5\times x^2\times x^8\right)+\dots \dots \dots \dots +\left(x^5\times x^{10}\times 1\right) \\ =11+11+11+11+11+11 \\ =66 \end{gathered}$$

Therefore, the correct answer is option (A).