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Question: There are 6 chocolates of Nestle, 8 of Cadbury, 10 of Perks, and 9 of Munch. The number of ways in w...

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

Explanation

Solution

We will use some general formula i.e. the number of solutions of n non-negative integral variables lets say x1,x2,x3,x4..........xn{x_1},{x_2},{x_3},{x_4}..........{x_n} and their sum is P i.e.,x1+x2+x3+x4+..........+xn=P{x_1} + {x_2} + {x_3} + {x_4} + .......... + {x_n} = P. Then the number of solutions of the above equation is P+n−1Cn−1{}^{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 x1,x2,x3,x4..........xn{x_1},{x_2},{x_3},{x_4}..........{x_n} and their sum is P ,
i.e.,x1+x2+x3+x4+..........+xn=P{x_1} + {x_2} + {x_3} + {x_4} + .......... + {x_n} = P
Then the number of solutions of the above equation is P+n−1Cn−1{}^{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=7N + 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+4−1C4−1−1{}^{7 + 4 - 1}{C_{4 - 1}} - 1
On simplifying we get,
=10C3−1= {}^{10}{C_3} - 1
Using,nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
⇒10C3−1=10!3!(10−3)!−1\Rightarrow {}^{10}{C_3} - 1 = \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}} - 1
=10!3!7!−1= \dfrac{{10!}}{{3!7!}} - 1
Using n!=n(n−1)!n! = n(n - 1)!
=10×9×8×7!3!7!−1= \dfrac{{10 \times 9 \times 8 \times 7!}}{{3!7!}} - 1
On cancelling common terms we get,
=10×9×83!−1= \dfrac{{10 \times 9 \times 8}}{{3!}} - 1
Expanding the value of 3!3!
=10×9×83×2×1−1= \dfrac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}} - 1
On simplifying we get,
=120−1= 120 - 1
=119= 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 nC2=n!2!(n−2)!{}^n{C_2} = \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}
Using n!=n(n−1)!n! = n(n - 1)!
⇒n!2!(n−2)!=n(n−1)(n−2)!2!(n−2)!\Rightarrow \dfrac{{n!}}{{2!\left( {n - 2} \right)!}} = \dfrac{{n(n - 1)(n - 2)!}}{{2!\left( {n - 2} \right)!}}
=n(n−1)2= \dfrac{{n(n - 1)}}{2}, we’ll be using the further solution
We know that
N+C+P+M=7N + C + P + M = 7,
When N=0
C+P+M=7C + P + M = 7, then the number of the solution will be 7+3−1C3−1{}^{7 + 3 - 1}{C_{3 - 1}}
=9C2= {}^9{C_2}
=9×82= \dfrac{{9 \times 8}}{2}
=36= 36

When N=1
C+P+M=6C + P + M = 6, then the number of the solution will be 6+3−1C3−1{}^{6 + 3 - 1}{C_{3 - 1}}
=8C2= {}^8{C_2}
=8×72= \dfrac{{8 \times 7}}{2}
=28= 28
When N=2
C+P+M=5C + P + M = 5, then the number of the solution will be 5+3−1C3−1{}^{5 + 3 - 1}{C_{3 - 1}}
=7C2= {}^7{C_2}
=7×62= \dfrac{{7 \times 6}}{2}
=21= 21
When N=3
C+P+M=4C + P + M = 4, then the number of the solution will be 4+3−1C3−1{}^{4 + 3 - 1}{C_{3 - 1}}
=6C2= {}^6{C_2}
=6×52= \dfrac{{6 \times 5}}{2}
=15= 15
When N=4
C+P+M=3C + P + M = 3, then the number of the solution will be 3+3−1C3−1{}^{3 + 3 - 1}{C_{3 - 1}}
=5C2= {}^5{C_2}
=5×42= \dfrac{{5 \times 4}}{2}
=10= 10
When N=5
C+P+M=2C + P + M = 2, then the number of the solution will be 2+3−1C3−1{}^{2 + 3 - 1}{C_{3 - 1}}
=4C2= {}^4{C_2}
=4×32= \dfrac{{4 \times 3}}{2}
=6= 6
When N=6
C+P+M=1C + P + M = 1, then the number of the solution will be 1+3−1C3−1{}^{1 + 3 - 1}{C_{3 - 1}}
=3C2= {}^3{C_2}
=3×22= \dfrac{{3 \times 2}}{2}
=3= 3
Therefore the total number of ways=36+28+21+15+10+6+3
=119