Question

10 Different toys are to be distributed among 10 children, find the total number of ways of distribution of these toys, so that exactly two children did not get any toy.

Answer 1

Here, we will consider two conditions for distributing the toys to the children by using the combination formula. We will then add the number of ways obtained in these conditions. Solving the equation further will give us the total number of ways of distribution.

Formula Used:
Combination is given by the formula ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$.

Complete step by step solution:
We are given that 10 Different toys are to be distributed among 10 children.
We are also given exactly two children who did not get any toy.
First, we will consider that exactly two children do not get any toy, one child gets three toys and the remaining all the other 7 children get one toy each.
So, the number of ways of distribution of 10 toys such that exactly two children do not get any toy, one child gets three toys and the remaining all the other 7 children get one toy each.
Now, by using the combination formula ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$, we get
${N_1} = \dfrac{{10!}}{{2!3!7!}} \times 10!$
First, we will consider that exactly two children do not get any toy, two children get two toys each and the remaining all the other 6 children get one toy each.
So, the number of ways of distribution of 10 toys such that exactly two children do not get any toy, two child gets two toys each and the remaining all the other 6 children get one toy each
Now, by using the combination formula ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$, we get
${N_2} = \dfrac{{10!}}{{2!2!2!2!6!}} \times 10!$
$\Rightarrow {N_2} = \dfrac{{10!}}{{{{\left( {2!} \right)}^4}6!}} \times 10!$
The total number of ways is the sum of both the conditions. Therefore, we get
$N = {N_1} + {N_2}$

By substituting the values of ${N_1}$ and ${N_2}$ , we get
$\Rightarrow N = \dfrac{{10!}}{{2!3!7!}} \times 10! + \dfrac{{10!}}{{{{\left( {2!} \right)}^4}6!}} \times 10!$
By taking the common term and by using the rule $n! = n \times \left( {n - 1} \right)!$ we get
$\Rightarrow N = \dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}\left( {\dfrac{1}{{3 \times 7}} + \dfrac{1}{{{{\left( {2!} \right)}^2}}}} \right)$
Simplifying the expression inside the bracket, we get
$\Rightarrow N = \dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}\left( {\dfrac{1}{{21}} + \dfrac{1}{4}} \right)$
Now taking LCM of the terms inside the bracket, we get
$\Rightarrow N = \dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}\left( {\dfrac{1}{{21}} \times \dfrac{4}{4} + \dfrac{1}{4} \times \dfrac{{21}}{{21}}} \right)$
$\Rightarrow N = \dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}\left( {\dfrac{{4 + 21}}{{82}}} \right)$
Adding the terms in the numerator, we get
$\Rightarrow N = \dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}\left( {\dfrac{{25}}{{82}}} \right)$
Therefore, the total number of ways of distribution of these toys, so that exactly two children did not get any toy is $\dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}\left( {\dfrac{{25}}{{82}}} \right)$.

Note:
We know that there is not much difference between permutation and combination. Permutation is the way of arranging numbers in some order whereas combination is the way of selecting items where order doesn’t matter. Both the combination and permutation is the way of arrangement. But, here we will not use permutation because the order of toys is not necessary.