Find the number of strings of $4$ letters that can be formed with the letters of the word $EXAMINATION$?
There are $11$ letters in the word $EXAMINATION$ of which there are $5$ distinct letters (E, X, M, T, O), $2$ A's, $2$ I's and $2$ N's.
All $4$ letters are distinct
$4$ distinct letters can be selected from $8$ distinct letters (E, X, M, T, O, A, I, N) in
${^{8}}{C}_{4} = \dfrac{8!}{4! \times 4!} = \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}= 70$ ways
Amongst themselves, the selected $4$ letters can be arranged in $4! = 24$ ways
$\therefore \;$ Number of possible arrangements where all $4$ letters are distinct $= 70 \times 24 = 1680$ ways
$1$ pair of letters, $2$ distinct letters
$1$ pair of letters can be selected from the $3$ pair's of letters in ${^{3}}{C}_{1} = 3$ ways
$2$ distinct letters can be selected from the remaining $7$ distinct letters in
${^{7}}{C}_{2} = \dfrac{7!}{5! \times 2!} = \dfrac{7 \times 6}{2 \times 1} = 21$ ways
Amongst themselves, the selected $4$ letters can be arranged in $\dfrac{4!}{2!} = 12$ ways
$\therefore \;$ Number of possible arrangements where $1$ pair of letters and $2$ letters distinct are selected $= 3 \times 21 \times 12 = 756$ ways
$2$ pairs of letters
$2$ pairs of letters can be selected from the $3$ pairs of letters in ${^{3}}{C}_{2} = 3$ ways
Amongst themselves, the selected $4$ letters can be arranged in $\dfrac{4!}{2! \times 2!} = \dfrac{4 \times 3}{2 \times 1} = 6$ ways
$\therefore \;$ Number of possible arrangements where $2$ pairs of letters are selected $= 3 \times 6 = 18$ ways
$\therefore \;$ Total number of possible strings $= 1680 + 756 + 18 = 2454$ ways