Verified Solution • 2025–26 Session moderate

QJEEM23Mae866: JEE Main 2023 Mathematics - Class 11 JEEM23Mae866 Solution: If the number of words, with or without meaning, which can be…

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 :

Multiple Choice Options

A 5670
B 1890
C 2835
D 945

Step-by-Step Verified Solution

Total number of words = \frac{11!}{2! 2! 2!}

Number of words in which

\mathrm{C}

and

\mathrm{S}

are together

= \frac{10!}{2! 2! 2!} \times 2!

So, required number of words

\begin{aligned} = \frac{11!}{2! 2! 2!} - \frac{10!}{2! 2!} \\ = \frac{11 \times 10!}{2! 2! 2!} - \frac{10!}{2! 2!} \\ = \frac{10!}{2! 2!} [\frac{11}{2} - 1] = \frac{10!}{2! 2!} \times \frac{9}{2} \\ = 5670 \times 6! \\ \Rightarrow k (6!) = 5670 \times 6! \\ \Rightarrow k = 5670 \end{aligned}

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