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# Show that only 286 four letter words can be formed out of the letter of the word ‘infinite’.  Verified
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Hint: We are going to use formulae of permutations and combinations with repetition of letters.

In the word ‘infinite’, we have a total 8 letters out of which are 3 i’s, 2 n’s and 3 different letters. 0
Number of different arrangements of ‘n’ letters where ${n_1}$ repeated letters, ${n_2}$ repeated letters ... ${n_k}$ repeated letters, is equal to $$\dfrac{{n!}}{{{n_1}!{n_2}!...{n_k}!}}$$.

From the word ‘infinite’, we are taking 4 letters to form all possible four lettered words.

(i). When we take 4 letters in which all are different, then
Number of words formed = $^5{C_4} \cdot 4! = 120$
[We have 5 unique letters in the word ‘infinite’, so we took $^5{C_4}$.]

(ii). When we take 2 alike and 2 different letters, then
Number of words formed = $$^2{C_1}{ \cdot ^4}{C_2} \cdot \dfrac{{4!}}{{2!}} = 144$$
[We have ‘n’ is repeated two times, ‘i’ is repeated three times, selecting two letters from these repeated letters is $$^2{C_1}$$]

(iii).When we take 2 alike and 2 alike letters, then
Number of words formed = $^2{C_2} \cdot \dfrac{{4!}}{{2!2!}} = 6$
[Taking two similar and two other similar letters from the given word is $^2{C_2}$.]

(iv).When we take 3 alike and 1 different letter, then
Number of words formed = $^1{C_1}{ \cdot ^4}{C_1} \cdot \dfrac{{4!}}{{3!}} = 16$

Total number of four digit letters = 120 + 144 + 6 +16 = 286.
Hence proved.

Note: In the given word ‘infinite’, we can observe this 8 lettered word, it actually consists of only 5 different digits i.e., (i, n, f, t, e). Two letters are repeated. So we used the formula of arrangement of n letters with repetition of letters.