Question

Find number of arrangements of 4 - letters taken from the word EXAMINATION.

Answer 1

EXAMINATION

$E\to 1$

$X\to 1$

$A=2$

$M\to 1$

$N\to 2$

$T\to 1$

$O\to 1$

$I\to 2$

Case $1$ : Same letter each of $2$ kinds $AANN,AAII,NNII$

each word can be arranged = $\frac{4!}{2!2!}$

Total cases = $3\times \frac{4!}{2!2!}=18$ ways

Case $2:2$ same letter of one kind

$AA$__

$NN$__

$II$__

Now consider the case $AA$__ for selecting the next two letter you can select next of $1N$ & $1I.$ So you need to select $2$ letters from $7$ left so, say for $AA$__ case to be

no. of permutation $={}^7{C}_2.\frac{4!}{2!}$

So total permutation $=3\times {}^7{C}_2.\frac{4!}{2!}=378$ way

Case $3:$ NO. repatition of any letter

Total no. of permutation ${=}^8{C}_4\times 4!=1680$ ways

So, total no of ways

$=1680+378+18=2076$ ways