Question

There are $13$ letters of $8$ different sorts $I,I,I,S,S,T,T,L,L,A,O,N,D$. In finding groups of $4$, how many permutations can be made if following are the possibilities to be considered? If $2$ are alike of one kind and $2$ are alike of other kind.

• A. $44$
• B. $52$
• C. $36$
• D. $102$

Answer 1

Answer: (C)

$36$

We need to form groups of $4$ letters so that $2$letters are of same kind and other $2$ letters are also of same kind but different from first letter.

Out of given letters there are $4$ letters which repeat , we need to choose $2$ out of them $={}^4{C}_2=6$

Now these $4$ letters can rearranged in $\frac{4!}{2!\times 2!}$ ways

$=\frac{24}{4}$

$=6$

$\Rightarrow$ Required $=6\times 6=36.$

Hence, the answer is $36.$