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Number of ways of distributing $ \left( {p + q + r} \right) $ different things to $ 3 $ persons so that one person gets p things, second one gets q things and third person gets r things is:
(A) $ \dfrac{{\left( {p + q + r} \right)!}}{{p!q!r!}} \times 3! $
(B) $ \dfrac{{\left( {p + q + r} \right)!}}{{p!q!r!}} $
(C) $ \dfrac{{\left( {p + q + r} \right)!}}{{3!p!q!r!}} $
(D) $ \dfrac{{\left( {p + q + r} \right)!}}{{3!pqr}} $

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Last updated date: 14th Jul 2024
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Answer
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Hint: The given question revolves around the concepts of permutations and combinations. According to the problem, we have $ \left( {p + q + r} \right) $ different things to be distributed among three different people such that one person gets p things, second person gets q things and third person gets r things.

Complete step by step solution:
So, we have a total of $ \left( {p + q + r} \right) $ different things to be distributed among three persons such that the first person gets p things.
So, number of ways to select p different things out of $ \left( {p + q + r} \right) $ different things is $ ^{\left( {p + q + r} \right)}{C_p} $ .
Then, the second person gets q different things.
So, the number of ways to select p different things out of remaining $ \left( {q + r} \right) $ different things is $ ^{\left( {q + r} \right)}{C_q} $ .
Also, third person gets different things.
So, the number of ways to select r different things out of remaining r different things is $ ^r{C_r} $ .
Also, we know that p, q and r are different. So, first person has to be assigned with p things. Similarly, the second person has to be assigned with q things and third person with r things. So, there are $ 3! $ ways of arranging three people.
Now, we use the fundamental principle of counting.
So, the total number of ways of distributing $ \left( {p + q + r} \right) $ different things among three different people are $ ^{\left( {p + q + r} \right)}{C_p}{ \times ^{\left( {q + r} \right)}}{C_q}{ \times ^r}{C_r} \times 3! $ .
Now, we expand the formulae of combinations.
 $ \Rightarrow \dfrac{{\left( {p + q + r} \right)!}}{{p!\left( {q + r} \right)!}} \times \dfrac{{\left( {q + r} \right)!}}{{q!r!}} \times \dfrac{{r!}}{{r!0!}} \times 3! $
Simplifying the expression further and substituting the value of $ 0! $ as one, we get,
 $ \Rightarrow \dfrac{{\left( {p + q + r} \right)!}}{{p!}} \times \dfrac{1}{{q!}} \times \dfrac{1}{{r!}} \times 3! $
 $ \Rightarrow \dfrac{{\left( {p + q + r} \right)!}}{{p!q!r!}} \times 3! $
So, the option (A) is correct.
So, the correct answer is “Option A”.

Note: One should know about the principle rule of counting or the multiplication rule. Care should be taken while handling the calculations. Calculations should be verified once so as to be sure of the answer.