If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which $\mathrm{C}$ and $\mathrm{S}$ do not come together, is $(6 !) \mathrm{k}$, then $\mathrm{k}$ is equal to :
Solution
$\text { Total number of words }=\frac{11 !}{2 ! 2 ! 2 !}$
<br/><br/>Number of words in which $\mathrm{C}$ and $\mathrm{S}$ are together
<br/><br/>$=\frac{10 !}{2 ! 2 ! 2 !} \times 2 \text { ! }$
<br/><br/>So, required number of words
<br/><br/>$$
\begin{aligned}
& =\frac{11 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\
& =\frac{11 \times 10 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\
& =\frac{10 !}{2 ! 2 !}\left[\frac{11}{2}-1\right]=\frac{10 !}{2 ! 2 !} \times \frac{9}{2} \\\\
& =5670 \times 6 ! \\\\
& \Rightarrow k(6 !)=5670 \times 6 ! \\\\
& \Rightarrow k=5670
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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