Question

# Ten different toys are to be distributed among ten children. What is the total number of ways of distributing these toys, so that exactly two children do not get any toys?(a). $\dfrac{{10!}}{{2!.3!.7!}}$(b). $\dfrac{{10!}}{{{{\left( {2!} \right)}^4} \times 6!}}$(c). ${\left( {10!} \right)^2}\left[ {\dfrac{1}{{{{\left( {2!} \right)}^4} \times 6!}} + \dfrac{1}{{2!{\text{ }} \times 3!}}} \right]$(d). $\dfrac{{10!{\text{ }} \times 10!}}{{{{\left( {2!} \right)}^2} \times 6!}}\left[ {\dfrac{{25}}{{84}}} \right]$

Verified
146.1k+ views
Hint: We get two cases when two out of ten children do not get any toys. Evaluate the two cases and add them to get the required answer.

We need to distribute 10 toys among ten children out of which 2 children do not get any toys.
So, we get two cases such that in the first case, two children do not get any toys, one child gets three toys and all the other children get one toy each. In the second case, again as required, two children do not get any toy, two children get two toys each and the remaining children get one toy each.
Hence, for case 1, we have the number of ways of distributing 10 different toys such that two children do not get any toys, one child gets three toys and all the other children get one toy each is given as follows:
${N_1} = \dfrac{{10!}}{{2!3!7!}} \times 10!.............(1)$
For case 2, we have the number of ways of distributing 10 different toys such that two children do not get any toy, two children get two toys each and the remaining children get one toy each is given as follows:
${N_2} = \dfrac{{10!}}{{{{\left( {2!} \right)}^4}6!}} \times 10!.............(2)$
The total number of ways is the sum of ${N_1}$ and ${N_2}$.
$N = {N_1} + {N_2}$
Substituting equation (1) and equation (2), we have:
$N = \dfrac{{10! \times 10!}}{{2!3!7!}} + \dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^4}6!}}$
Taking $\dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}$ as a common term, we get:
$N = \dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}\left[ {\dfrac{1}{{21}} + \dfrac{1}{4}} \right]$
$N = \dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}\left[ {\dfrac{{4 + 21}}{{84}}} \right]$
$N = \dfrac{{10! \times 10!}}{{{{\left( {2!} \right)}^2}6!}}\left[ {\dfrac{{25}}{{84}}} \right]$
Hence, the correct answer is option (d).

Note: Do not attempt to directly distribute 10 different toys among 8 children because any two children among the 10 can be left out and there is a combination in choosing the 8 children itself.