Question

Find the number of ways of distributing n identical objects among n persons if at least n – 3 persons get none of these objects.

Answer 1

Hint : Here, all n objects are identical so here only how many objects are given to a person will matter but not which object is given to a person will matter. Also, here the number of objects is equal to the total number of people that is n and also we have to distribute objects in such a way that at least n – 3 persons get none of these objects. So, we can say that we have to distribute objects such that at most 3 people get the object

** Complete step-by-step answer** :
Now, number of ways to distribute n objects to exactly three people will be given as,
Firstly, we have to choose three persons out of n persons which is equals to ${}^{n}{{C}_{3}}\cdot$
Secondly, we have to find the number of ways questions being distributed among them whose sum will be equals to n which is equals to ${}^{n-1}{{C}_{2}}$
So, number of ways to distribute n objects to exactly 3 people = ${}^{n}{{C}_{3}}\cdot {}^{n-1}{{C}_{2}}$ ……( i )
Now, number of ways to distribute n objects to exactly two people will be given as,
Firstly, we have to choose two persons out of n persons which is equals to ${}^{n}{{C}_{2}}\cdot$
Secondly, we have to find the number of ways questions being distributed among these two persons, whose sum will be equals to n objects which is equals to ${}^{n-1}{{C}_{1}}$
So, number of ways to distribute n objects to exactly two people = ${}^{n}{{C}_{2}}\cdot {}^{n-1}{{C}_{1}}$ …..( ii )
Now, number of ways to distribute n objects to exactly one person will be given as,
Firstly, we have to choose one person out of n persons which is equals to ${}^{n}{{C}_{1}}$ .
Secondly, we have to find the number of ways questions being distributed among this one person, whose sum of subjects will be equals to n objects which is equals to ${}^{n-1}{{C}_{0}}$
So, number of ways to distribute n objects to exactly two people = ${}^{n}{{C}_{1}}\cdot {}^{n-1}{{C}_{0}}$ ……( iii )
Then, the number of ways of distributing n identical objects among n persons if at least n – 3 persons get none of these objects will be given as sum of ( i ), ( ii ), ( iii )
${}^{n}{{C}_{3}}\cdot {}^{n-1}{{C}_{2}}+{}^{n}{{C}_{2}}\cdot {}^{n-1}{{C}_{1}}+{}^{n}{{C}_{1}}\cdot {}^{n-1}{{C}_{0}}$
Now, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ where $n!=n\cdot (n-1)\cdot (n-2)\cdot ......\cdot 3\cdot 2\cdot 1$
So, $\dfrac{n!}{3!\left( n-3 \right)!}\cdot \dfrac{(n-1)!}{2!\left( n-1-2 \right)!}+\dfrac{n!}{2!\left( n-2 \right)!}\cdot \dfrac{(n-1)!}{1!\left( n-1-1 \right)!}+\dfrac{n!}{1!\left( n-1 \right)!}\cdot \dfrac{(n-1)!}{0!\left( n-1-0 \right)!}$
= $\dfrac{n(n-1)(n-2)}{3\cdot 2\cdot 1}\cdot \dfrac{(n-1)\cdot (n-2)}{2\cdot 1}+\dfrac{n\cdot (n-1)}{2\cdot 1}\cdot (n-1)+n$
=\dfrac{n{{(n-1)}^{2}}}{2}\left\\{ \dfrac{{{(n-2)}^{2}}}{6}+1 \right\\}+n
= \dfrac{n{{(n-1)}^{2}}}{2}\left\\{ \dfrac{({{n}^{2}}-4n+10}{6} \right\\}+n
Hence, the number of ways of distributing n identical objects among n persons if at least n – 3 persons get none of these objects are \dfrac{n{{(n-1)}^{2}}}{2}\left\\{ \dfrac{({{n}^{2}}-4n+10)}{6} \right\\}+n.

Note : While solving combination problems always remember for each parameter we need to find a number of ways individually. Always remember the formula of combination when we need to choose r items from total n items. Calculation should be accurate as expression can be lengthy and complex too.