Question

How many different permutations can be formed from the letter of the word EXAMINATION taken four at a time?

Answer 1

We can choose 4 letters from the 11 listed in part as under.
(i) All the four different:
We have 8 different types of letter and out of these 4 can be arranged in
${}^8{P}_4=\frac{8!}{4!}=8\times 7\times 6\times 5=1680$
or 4 can be selected in
${}^8{C}_4$ ways are arranged in
${}^8{C}_4\times 4!=\frac{8!}{4!4!}4!=\frac{8!}{4!}=1680$ .....(1)
(ii) Two different and two alike.
We have 3 pairs of like letters out of which one pair can be choosen in ${}^3{C}_1=3$ ways. Now we have to choose two out of the remaining 7 defferent types of letter which can be done in
${}^7{C}_2=\frac{7!}{5!2!}=\frac{7\times 6}{2}=21$ ways.
Hence the total number of groups of 4 letter in which 2 are different and 2 are alike is $3\times 21=63$groups
Each group has 4 letters out of which 2 are alike and they can be arranged amongst themselves in $\frac{4!}{2!}=12$ ways. Hence the total number of words is
$63\times 12=756.$ ....(2)
(iii) Two alike of one kind two alike of other kind.
Out of 3 pairs of like letters we can choose 2 pairs in
${}^3{C}_2$ways $=3$ ways.
One such group is $MM$ $AA$.
These four letter out of which 2 are alike of one kind and 2 alike of other kind, can be arranged in $\frac{4!}{2!2!}=6$ways.
Hence the total number of words of this type is $3\times 6=18$ .....(3)
Therefore from (i), (ii) and (iii) the number of 4 letter word is
$1680+756+18=2454$.
By (1), (2) and (3)