Question

There are 6 chocolates of Nestle, 8 of Cadbury, 10 of Perk and 9 of Munch. The number of ways in which a child can buy 7 chocolates is

Answer 1

We will write the given statement in the form of an equation. The number of solutions to that equation will be our answer to the problem. We will use the concept of combinations to find the number of solutions to the equation. Then we will subtract 1 from the number of solutions to adjust for the less number of chocolates of Nestle.
Formulas used:
We will use the following formulas:
We know that number of solutions of the equation ${x_1} + {x_2} + {x_3} + ... + {x_n} = S$ is 1.${}^{S + n - 1}{C_{n - 1}}$ where ${x_1},{x_2},{x_3},...,{x_n}$ are non-negative integers.
2.${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ , where $n$ is the total number of objects and $r$ is the number of objects to be selected.
3.$n! = n\left( {n - 1} \right)\left( {n - 2} \right)...3 \cdot 2 \cdot 1$

Complete step-by-step answer:
Let the number of chocolates of Cadbury be represented by C, the number of chocolates of Perk be represented by P, the number of chocolates of Nestle be represented by N and the number of chocolates of Munch be represented by M.
We know that a total of 7 chocolates have to be selected. So:
$N + C + P + M = 7$
We will find the number of solutions to this equation. We will substitute 7 for $S$ and 4 for $n$ in the formula ${}^{S + n - 1}{C_{n - 1}}$:
$\Rightarrow$ ${}^{7 + 4 - 1}{C_{4 - 1}} = {}^{10}{C_3}$
The number of ways in which the child can select the chocolates will be 1 less than the number of solutions of the above equation. This is because there are only 6 chocolates of Nestle and the case where 7 Nestle chocolates are selected is not possible.
We will find the number of ways in which a child can buy 7 chocolates:
$\Rightarrow$ ${}^{10}{C_3} - 1$
We will substitute 10 for $n$ and 3 for $r$ in the formula ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$:
$\Rightarrow$ ${}^{10}{C_3} - 1 = \dfrac{{10!}}{{\left( {10 - 3} \right)!3!}} - 1 = \dfrac{{10 \times 9 \times 8 \times 7!}}{{7! \times 6}} - 1$
Simplifying the terms, we get
$\Rightarrow {}^{10}{C_3} - 1 = 120 - 1 = 119$
$\therefore$ There are 119 ways in which a child can choose 7 chocolates from 6 chocolates of Nestle, 8 of Cadbury, 10 of Perk and 9 of Munch.

Note: We must know that the formula for combination is used in cases where the order of selection does not matter whereas in cases where the order of selection also matters, we should use the formula for permutations. The formula for permutations is given by ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ where $n$ is the total number of objects and $r$ is the number of objects to be selected. Permutation and combination both are useful in finding probability.