If the number of words,with or without meanin | vedclass

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(D) The word $MATHEMATICS$ has $11$ letters: $M, M, A, A, T, T, H, E, I, C, S$.Total arrangements = $\frac{11!}{2! 2! 2!} = \frac{39916800}{8} = 49896

Vedclass · Accepted Answer

$5670$

Vedclass · Answer

(D) The word $MATHEMATICS$ has $11$ letters: $M, M, A, A, T, T, H, E, I, C, S$.Total arrangements = $\frac{11!}{2! 2! 2!} = \frac{39916800}{8} = 4989600$.To find the number of arrangements where $C$ and $S$ are together,treat $(CS)$ as one unit. Now we have $10$ units: $M, M, A, A, T, T, H, E, I, (CS)$.Arrangements with $C$ and $S$ together = $\frac{10!}{2! 2! 2!} 	imes 2! = \frac{3628800}{8} 	imes 2 = 907200$.Number of words where $C$ and $S$ do not come together = $4989600 - 907200 = 4082400$.We are given that this is $(6!) k = 720k$.$720k = 4082400$.$k = \frac{4082400}{720} = 5670$.

If the number of words,with or without meaning,which can be made using all the letters of the word $MATHEMATICS$ in which $C$ and $S$ do not come together,is $(6 !) k$,then $k$ is equal to $............$.

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