Question

In how many ways fifteen different items may be given to A, B, C such that $A$ gets $3$, $B$ gets $5$ and remaining goes to C.

Answer 1

Hint: In order to solve this question we have to apply the concept of combination since all the items are different. We will find the possible number of ways individually for all the A,B and C.

Formula used: ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$

Complete step-by-step solution:
It is stated in the question that there are $15$ different items where $A$ gets $3$, $B$ gets $5$ and therefore $C$ gets $15 - 3 - 5 = 7$.
Since $A$ get $3$ out of $15$ different items therefore we can write-
${}^{15}{C_3} = \dfrac{{15!}}{{3!(15 - 3)!}} = \dfrac{{15!}}{{3! \times 9!}}$
Again, $B$ gets $5$ out of $15$ different items therefore we can write-
${}^{15}{C_5} = \dfrac{{15!}}{{5!(15 - 5)!}} = \dfrac{{15!}}{{5! \times 10!}}$
Now, $C$ gets $7$ out of $15$ different items therefore we can write-
${}^{15}{C_7} = \dfrac{{15!}}{{7!(15 - 7)!}} = \dfrac{{15!}}{{7! \times 8!}}$
Therefore, the total number of ways in which we can arrange it $ = {}^{15}{C_3} \times {}^{15}{C_5} \times {}^{15}{C_7}$$ = \dfrac{{15!}}{{3! \times 9!}} \times \dfrac{{15!}}{{5! \times 10!}} \times \dfrac{{15!}}{{7! \times 8!}}$
$ = \dfrac{{15! \times 5!}}{{3! \times 7!}}$

Thus, the required number of ways in which fifteen different items may be given to A, B, C such that $A$ gets $3$, $B$ gets $5$ and the remaining goes to $C = \dfrac{{15! \times 5!}}{{3! \times 7!}}$

Note: Permutation can be defined as the technique of arranging all the members in a group. In other ways, we can say that if the set has already been ordered then re-ordering the elements is called the process of permuting. It is useful almost in every part of mathematics. It is basically used for arranging people, digits, numbers etc.
The basic formula for permutation ${}^n{p_r} = \dfrac{{n!}}{{(n - r)!}}$
Combination can be regarded as a way of choosing items from a group such that the order of selection does not matter. It is basically used for selecting menu, food, clothes, balls and many more.
The basic formula for combination ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$