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Question: Number of combinations of p letters from 3p letters of which p are l, p are m and rest are distinct,...

Number of combinations of p letters from 3p letters of which p are l, p are m and rest are distinct, are given by_____. Fill in the blank.
A. (p+1)2p−1\left( {p + 1} \right){2^{p - 1}}
B. p2pp^{2p}
C. (p+1)2p\left( {p + 1} \right){2^p}
D. (p+2)2p−1\left( {p + 2} \right){2^{p - 1}}

Explanation

Solution

This question is based on the concept of permutation and combination, if the set is already ordered, then the rearranging of its elements is called the process of permuting and the combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter

Formula used:
nPr = n!/(n−r)! and nCr = n!/[r! (n−r)!]nPr{\text{ }} = {\text{ }}n!/\left( {n - r} \right)!\,and\,nCr{\text{ }} = {\text{ }}n!/\left[ {r!{\text{ }}\left( {n - r} \right)!} \right]

Complete step by step answer:
From p m’s, we can have any of 0,1,2…….pm’s. From p l’s we can have any of 0,1,…..p l’s. Besides there for remaining p letters we have only two combinations namely 0 or 1. So the required number of combinations is =coefficient of xp{x^p} in
(1+x+x2+x3+.....+xp)(1+x+x2+....+xp)(1+x)(1+x).....(1+x)p−factors\left( {1 + x + {x^2} + {x^3} + ..... + {x^p}} \right)\left( {1 + x + {x^2} + .... + {x^p}} \right)\dfrac{{\left( {1 + x} \right)\left( {1 + x} \right).....\left( {1 + x} \right)}}{{p - factors}}
Coefficient of xp{x^p} in (1−xp+1)2(1−x)2(1+x)p\dfrac{{{{\left( {1 - {x^{p + 1}}} \right)}^2}}}{{{{\left( {1 - x} \right)}^2}}}{\left( {1 + x} \right)^p}

Coefficient of xp{x^p}in (1−xp+1)(1−x)−2\left( {1 - {x^{p + 1}}} \right){\left( {1 - x} \right)^{ - 2}}.
Coefficient of xp{x^p} in (1+x)p(1−x)−2{\left( {1 + x} \right)^p}{\left( {1 - x} \right)^{ - 2}}.
As (1−xp+1)2{\left( {1 - {x^{p + 1}}} \right)^2}does not contain xp{x^p}.
Therefore, coefficient of xp{x^p} in [2−(1−x)]p(1−x)−2{\left[ {2 - \left( {1 - x} \right)} \right]^p}{\left( {1 - x} \right)^{ - 2}}.
Coefficient of xp{x^p} in 2p(1−x)−2−p2p−1(1−x)−1+pC22r−2(1−x)0p2p{\left( {1 - x} \right)^{ - 2}} - p{2^{p - 1}}{\left( {1 - x} \right)^{ - 1}} + p{C_2}{2^{r - 2}}{\left( {1 - x} \right)^0}\\_p.
C32p−3(1−x)+...+(−1)p∙pCp(1−x)p−2{C_3}{2^{p - 3}}\left( {1 - x} \right) + ... + \left( { - 1} \right){}^{p \bullet p}{C_p}{\left( {1 - x} \right)^{p - 2}}.
Hence coefficient of xp{x^p} in 2p(1−x)−2−p.2p−1(1−x)−1{2^p}{\left( {1 - x} \right)^{ - 2}} - p{.2^{p - 1}}{\left( {1 - x} \right)^{ - 1}}.
That is 2p+p2p−1{2^p} + p{2^{p - 1}}.

Therefore, option D is the correct answer.

Note: Alternative approach is that out of remaining 24 alphabets, remaining ‘P’ alphabets can be chosen in 24CP{}^{24}{C_P} ways, the number of ways of arranging 3p letters, in which ‘p’ are of one type and ‘P’ are of other type and rest P are distinct and equal to 24CP⋅3P!P!⋅P!{}^{24}{C_P} \cdot \dfrac{{3P!}}{{P! \cdot P!}}.