JEE MAIN - Mathematics (2023 - 8th April Evening Shift - No. 9)
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 :
5670
1890
2835
945
Explanation
$$
\text { Total number of words }=\frac{11 !}{2 ! 2 ! 2 !}
$$
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
$$ \begin{aligned} & =\frac{11 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\ & =\frac{11 \times 10 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\ & =\frac{10 !}{2 ! 2 !}\left[\frac{11}{2}-1\right]=\frac{10 !}{2 ! 2 !} \times \frac{9}{2} \\\\ & =5670 \times 6 ! \\\\ & \Rightarrow k(6 !)=5670 \times 6 ! \\\\ & \Rightarrow k=5670 \end{aligned} $$
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
$$ \begin{aligned} & =\frac{11 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\ & =\frac{11 \times 10 !}{2 ! 2 ! 2 !}-\frac{10 !}{2 ! 2 !} \\\\ & =\frac{10 !}{2 ! 2 !}\left[\frac{11}{2}-1\right]=\frac{10 !}{2 ! 2 !} \times \frac{9}{2} \\\\ & =5670 \times 6 ! \\\\ & \Rightarrow k(6 !)=5670 \times 6 ! \\\\ & \Rightarrow k=5670 \end{aligned} $$
Comments (0)
