If the number of words, with or without meaning, which can be made using all the letters o…

Question

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 :

Accepted Answer

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

Number of words in which $\mathrm{C}$ and $\mathrm{S}$ are together

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

So, required number of words

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

Solutions · 1