Question

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. \[\left( {p + 1} \right){2^{p - 1}}\]
B. $p^{2p}$
C. \[\left( {p + 1} \right){2^p}\]
D. \[\left( {p + 2} \right){2^{p - 1}}\]

Answer 1

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

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 \[{x^p}\] in
\[\left( {1 + x + {x^2} + {x^3} + ..... + {x^p}} \right)\left( {1 + x + {x^2} + .... + {x^p}} \right)\dfrac{{\left( {1 + x} \right)\left( {1 + x} \right).....\left( {1 + x} \right)}}{{p - factors}}\]
Coefficient of \[{x^p}\] in \[\dfrac{{{{\left( {1 - {x^{p + 1}}} \right)}^2}}}{{{{\left( {1 - x} \right)}^2}}}{\left( {1 + x} \right)^p}\]

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

Therefore, option D is the correct answer.

Note: Alternative approach is that out of remaining 24 alphabets, remaining ‘P’ alphabets can be chosen in \[{}^{24}{C_P}\] 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 \[{}^{24}{C_P} \cdot \dfrac{{3P!}}{{P! \cdot P!}}\].