Question

The number of ways of selecting 6 letters from 18 letters of which 6 are $x,6$ are y and rest are different is

• A. $2^5$
• B. $6.2^6$
• C. $8.2^5$
• D. None of these

Answer 1

The correct option is C $8.2^5$

We have $6x^{\prime }s,6y^{\prime }s$ and rest $6$ different.
$6$ letters can be selected in following ways :
All alike $=\left(\begin{array}{c}2\\ 1\end{array}\right)=2$
5 alike,1 different=$\left(\begin{array}{c}2\\ 1\end{array}\right)\times \left(\begin{array}{c}7\\ 1\end{array}\right)=14$
4 alike,2 different=$\left(\begin{array}{c}2\\ 1\end{array}\right)\times \left(\begin{array}{c}7\\ 2\end{array}\right)=42$
4 alike,2 other alike=$\left(\begin{array}{c}2\\ 2\end{array}\right)\times 2=2$
3 alike,3 different=$\left(\begin{array}{c}2\\ 1\end{array}\right)\times \left(\begin{array}{c}7\\ 3\end{array}\right)=70$
3 alike,3 other alike=$\left(\begin{array}{c}2\\ 2\end{array}\right)=1$
2 alike,4 different=$\left(\begin{array}{c}2\\ 1\end{array}\right)\times \left(\begin{array}{c}7\\ 4\end{array}\right)=70$
2 alike,3 other alike,1 different=$\left(\begin{array}{c}2\\ 2\end{array}\right)\times 2\times \left(\begin{array}{c}6\\ 1\end{array}\right)=12$
2 alike,2 other alike,2 different=$\left(\begin{array}{c}2\\ 2\end{array}\right)\times \left(\begin{array}{c}6\\ 2\end{array}\right)=15$
All different =$\left(\begin{array}{c}8\\ 6\end{array}\right)=28$
Therefore, the total number of ways$=(2+14+42+2+70+1+70+12+15+28)=256=$ $2^8=8\times 2^5$
Hence, option 'C' is correct.