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(C) The word $'SAHARANPUR'$ contains $10$ letters: $S: 1, A: 3, H: 1, R: 2, N: 1, P: 1, U: 1$.We need to form $3$ letter words. The possible cases are

Vedclass · Accepted Answer

$247$

Vedclass · Answer

(C) The word $'SAHARANPUR'$ contains $10$ letters: $S: 1, A: 3, H: 1, R: 2, N: 1, P: 1, U: 1$.We need to form $3$ letter words. The possible cases are:Case $1$: All $3$ letters are the same.We have only one letter '$A$' that appears $3$ times. Number of ways = $\binom{1}{1} 	imes \frac{3!}{3!} = 1 	imes 1 = 1$.Case $2$: All $3$ letters are different.There are $7$ distinct letters available $(S, A, H, R, N, P, U)$. Number of ways = $\binom{7}{3} 	imes 3! = 35 	imes 6 = 210$.Case $3$: $2$ letters are alike and $1$ is different.There are $2$ letters that appear at least twice ($A$ and $R$).If we choose '$A$' as the pair: $\binom{1}{1} 	imes \binom{6}{1} 	imes \frac{3!}{2!} = 1 	imes 6 	imes 3 = 18$.If we choose '$R$' as the pair: $\binom{1}{1} 	imes \binom{6}{1} 	imes \frac{3!}{2!} = 1 	imes 6 	imes 3 = 18$.Total for this case = $18 + 18 = 36$.Total number of words = $1 + 210 + 36 = 247$.

Total number of $3$ letter words that can be formed from the letters of the word $'SAHARANPUR'$ is equal to

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