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Question: In how many ways fifteen different items may be given to A, B, C such that \(A\) gets \(3\), \(B\) g...

In how many ways fifteen different items may be given to A, B, C such that AA gets 33, BB gets 55 and remaining goes to C.

Explanation

Solution

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: nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}

Complete step-by-step solution:
It is stated in the question that there are 1515 different items where AA gets 33, BB gets 55 and therefore CC gets 15−3−5=715 - 3 - 5 = 7.
Since AA get 33 out of 1515 different items therefore we can write-
15C3=15!3!(15−3)!=15!3!×9!{}^{15}{C_3} = \dfrac{{15!}}{{3!(15 - 3)!}} = \dfrac{{15!}}{{3! \times 9!}}
Again, BB gets 55 out of 1515 different items therefore we can write-
15C5=15!5!(15−5)!=15!5!×10!{}^{15}{C_5} = \dfrac{{15!}}{{5!(15 - 5)!}} = \dfrac{{15!}}{{5! \times 10!}}
Now, CC gets 77 out of 1515 different items therefore we can write-
15C7=15!7!(15−7)!=15!7!×8!{}^{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 =15C3×15C5×15C7 = {}^{15}{C_3} \times {}^{15}{C_5} \times {}^{15}{C_7} =15!3!×9!×15!5!×10!×15!7!×8! = \dfrac{{15!}}{{3! \times 9!}} \times \dfrac{{15!}}{{5! \times 10!}} \times \dfrac{{15!}}{{7! \times 8!}}
=15!×5!3!×7!= \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 AA gets 33, BB gets 55 and the remaining goes to C=15!×5!3!×7!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 npr=n!(n−r)!{}^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 nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}