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Question: Six apples and six mangoes are to be distributed among ten boys so that each boy receives at least o...

Six apples and six mangoes are to be distributed among ten boys so that each boy receives at least one fruit. Find the number of ways in which the fruits can be distributed.

Explanation

Solution

First of all we will assume the ten boys to be variables like x1,x2,x3{{x}_{1}},{{x}_{2}},{{x}_{3}} and so on. Now we have a total of 12 fruits and we have to distribute it among the ten boys in such a way that each boy receives at least one fruit from the 12 fruits and then we will distribute the remaining fruits. So, here we will use the formula to distribute ‘n’ things among ‘r’ groups, which is given as, n+r−1Cr−1^{n+r-1}{{C}_{r-1}} and also we will use the formula of combination, that is, nCrn!r!(n−r)!^{n}{{C}_{r}}\dfrac{n!}{r!\left( n-r \right)!} to solve this question.

Complete step-by-step answer:
We have been given a total of 12 fruits which we have to distribute among ten boys such that each boy receives at least one fruit.
Let us assume the boys to be x1,x2,x3,...x10{{x}_{1}},{{x}_{2}},{{x}_{3}},...{{x}_{10}} .
Since each boy must have at least one fruit, we will distribute the one fruit to each of the ten boys which means that we have distributed a total of 10 fruits among ten boys, i.e. one to each.
Now, the number of remaining fruits =12−10=2=12-10=2
So we have 2 fruits remaining with us after giving one fruit to each of the ten boys.
Now we will distribute these 2 fruits among ten boys.
Since we know that the distribution of ’n’ things among ‘r’ groups is given by as follows:
⇒\Rightarrow Number of ways of distribution =n+r−1Cr−1{{=}^{n+r-1}}{{C}_{r-1}}
So we have n=2 and r=10.
⇒\Rightarrow Number of ways of distribution =2+10−1C10−1=11C9{{=}^{2+10-1}}{{C}_{10-1}}{{=}^{11}}{{C}_{9}}
We know that =nCr=n!(n−r)!×r!{{=}^{n}}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}
⇒\Rightarrow Number of ways of distribution =11!(11−9)!×9!=9!×10×112!×9!=10×112=55=\dfrac{11!}{\left( 11-9 \right)!\times 9!}=\dfrac{9!\times 10\times 11}{2!\times 9!}=\dfrac{10\times 11}{2}=55
Therefore the number of ways the fruits can be distributed equals to 55.

Note: Be careful while using the formula n+r−1Cr−1^{n+r-1}{{C}_{r-1}}, sometimes by mistake we take n=10 and r=2 which is wrong and we get the incorrect answer. Assuming a variable for ten boys is not necessary but you will understand the question very clearly if you will do so and it is better to assume it variables.