Question

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

Answer 1

Hint: We will use some general formula i.e. the number of solutions of n non-negative integral variables lets say ${x_1},{x_2},{x_3},{x_4}..........{x_n}$ and their sum is P i.e.,${x_1} + {x_2} + {x_3} + {x_4} + .......... + {x_n} = P$. Then the number of solutions of the above equation is ${}^{P + n - 1}{C_{n - 1}}$. Then we’ll exclude those cases that will be contradicting the given data.

Complete step by step answer:

Given data:
Number of Nestle(N)=6
Number of Cadbury(C)=8
Number of Perk(P)=10
Number of Munch(M)=9
Now, we know that if there are n non-negative integral variables lets say ${x_1},{x_2},{x_3},{x_4}..........{x_n}$ and their sum is P ,
i.e.,${x_1} + {x_2} + {x_3} + {x_4} + .......... + {x_n} = P$
Then the number of solutions of the above equation is ${}^{P + n - 1}{C_{n - 1}}$ .
Now, we have to find the number of ways that the child can select 7 chocolates out of the given number of distinct chocolates.
We can write it as
$N + C + P + M = 7$
Where N, C, P, M are non-negative integers, but there N=6 therefore cannot select all the 7 nestle chocolates which means we have to subtract one solution out of the total number of solutions.
Then the number of ways that the child can select 7 chocolates will be ${}^{7 + 4 - 1}{C_{4 - 1}} - 1$
On simplifying we get,
$ = {}^{10}{C_3} - 1$
Using,${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$ \Rightarrow {}^{10}{C_3} - 1 = \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}} - 1$
$ = \dfrac{{10!}}{{3!7!}} - 1$
Using $n! = n(n - 1)!$
$ = \dfrac{{10 \times 9 \times 8 \times 7!}}{{3!7!}} - 1$
On cancelling common terms we get,
$ = \dfrac{{10 \times 9 \times 8}}{{3!}} - 1$
Expanding the value of \[3!\]
$ = \dfrac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}} - 1$
On simplifying we get,
$ = 120 - 1$
$ = 119$
Hence, The number of ways in which a child can buy 7 chocolates is 119.

Note: Alternative method of this solution can be
First, we find the value of ${}^n{C_2} = \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}$
Using $n! = n(n - 1)!$
$ \Rightarrow \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = \dfrac{{n(n - 1)(n - 2)!}}{{2!\left( {n - 2} \right)!}}$
$ = \dfrac{{n(n - 1)}}{2}$, we’ll be using the further solution
We know that
$N + C + P + M = 7$,
When N=0
$C + P + M = 7$, then the number of the solution will be ${}^{7 + 3 - 1}{C_{3 - 1}}$
$ = {}^9{C_2}$
$ = \dfrac{{9 \times 8}}{2}$
$ = 36$

When N=1
$C + P + M = 6$, then the number of the solution will be ${}^{6 + 3 - 1}{C_{3 - 1}}$
$ = {}^8{C_2}$
$ = \dfrac{{8 \times 7}}{2}$
$ = 28$
When N=2
$C + P + M = 5$, then the number of the solution will be ${}^{5 + 3 - 1}{C_{3 - 1}}$
$ = {}^7{C_2}$
$ = \dfrac{{7 \times 6}}{2}$
$ = 21$
When N=3
$C + P + M = 4$, then the number of the solution will be ${}^{4 + 3 - 1}{C_{3 - 1}}$
$ = {}^6{C_2}$
$ = \dfrac{{6 \times 5}}{2}$
$ = 15$
When N=4
$C + P + M = 3$, then the number of the solution will be ${}^{3 + 3 - 1}{C_{3 - 1}}$
$ = {}^5{C_2}$
$ = \dfrac{{5 \times 4}}{2}$
$ = 10$
When N=5
$C + P + M = 2$, then the number of the solution will be ${}^{2 + 3 - 1}{C_{3 - 1}}$
$ = {}^4{C_2}$
$ = \dfrac{{4 \times 3}}{2}$
$ = 6$
When N=6
$C + P + M = 1$, then the number of the solution will be ${}^{1 + 3 - 1}{C_{3 - 1}}$
$ = {}^3{C_2}$
$ = \dfrac{{3 \times 2}}{2}$
$ = 3$
Therefore the total number of ways=36+28+21+15+10+6+3
=119