Question

How many (i) combination (ii) permutation of $4$ letters be made from letters of the word $EXAMINATION$?

Answer 1

Given that,

$E,X,A,M,I,N,A,T,I,O,N$

Number of letters $=11$

$A$ occurs twice $N$ occurs twice $I$ occurs twice

Different letters are $E,X,A,M,I,N,T,O=8$

Case(1)

When two letters are Identical and remaining two are different letters used are

(i) Two A' and two out of $E.X.M.I.N$

$\begin{array}{cc}& no.ofarrangement\\ 1{\times }^7{c}_2& 1{\times }^7{c}_2\times \frac{4!}{2!}=252\end{array}$

(ii) Two $M^{{}^{\prime }s}$ and two out $E,X,M,I,A,T,O$

$\begin{array}{cc}& \\ 1{\times }^7{c}_2& |1{\times }^7{c}_2\times \frac{4!}{2!}=252\end{array}$

(iii) Two $I^{{}^{\prime }s}$ and out and $E.X.M.N.A.T.O$

$\begin{array}{cc}& \\ 1{\times }^7{c}_2& |1{\times }^7{c}_2\times \frac{4!}{2!}=252\\ & total=756\end{array}$

Case(II)

When two letters are identical and remaining two are identical

$\begin{array}{c}\begin{array}{ccc}& No.ofselecting& No.ofarragement\\ TwoA{}^{\prime s},twoN{}^{\prime s}& 1\times 1& 1\times 1\times \frac{4!}{2!2!}=6\\ TwoA{}^{\prime s},twoI{}^{\prime s}& 1\times 1& 1\times 1\times \frac{4!}{2!2!}=6\end{array}\\ TwoN{}^{\prime s},twoI{}^{\prime s}1\times 11\times 1\times \frac{4!}{2!2!}=6\\ total=18\end{array}$

Case(III)

$\begin{array}{ccc}& No.ofselections& No.ofarrangement\\ Fouroutof& {}^8{c}_4& {}^8{c}_4.4!=1680\\ E,X,A,M,I,O,N& & \end{array}$

Required Number $=756+18+1680=2454$

(ii) Required Number of ways $=\begin{array}{c}{}^{11}{P}_4\\ =\frac{11!}{4!}\\ =\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4}{4!}\\ =1663200\end{array}$