The number of 6-letter words,with or without | vedclass

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Question

(D) The word $	ext{MATHS}$ consists of $5$ distinct letters: $\{M, A, T, H, S\}$. We need to form a $6$-letter word such that each letter used appear

Vedclass · Accepted Answer

$1400$

Vedclass · Answer

(D) The word $	ext{MATHS}$ consists of $5$ distinct letters: $\{M, A, T, H, S\}$. We need to form a $6$-letter word such that each letter used appears at least twice.
Case $1$: Using $3$ distinct letters,each appearing twice.
Number of ways to choose $3$ letters from $5$ is $^5C_3 = 10$.
Number of arrangements of $6$ letters where each appears twice is $\frac{6!}{2!2!2!} = 90$.
Total words $= 10 	imes 90 = 900$.
Case $2$: Using $2$ distinct letters,one appearing $4$ times and one appearing $2$ times.
Number of ways to choose $2$ letters from $5$ is $^5C_2 = 10$.
Number of arrangements is $\frac{6!}{4!2!} 	imes 2 = 15 	imes 2 = 30$.
Total words $= 10 	imes 30 = 300$.
Case $3$: Using $2$ distinct letters,each appearing $3$ times.
Number of ways to choose $2$ letters from $5$ is $^5C_2 = 10$.
Number of arrangements is $\frac{6!}{3!3!} = 20$.
Total words $= 10 	imes 20 = 200$.
Total number of words $= 900 + 300 + 200 = 1400$.

The number of $6$-letter words,with or without meaning,that can be formed using the letters of the word $\text{MATHS}$ such that any letter that appears in the word must appear at least twice,is $...$

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