$\begin{array}{1 1}(A)\;151200\\(B)\;150200\\(C)\;153200\\(D)\;154200\end{array} $

- $C(n,r)=\large\frac{n!}{r!(n-r)!}$

Total no of consonants =6

Total no of vowels =6

Total no of letters in the word INTERMEDIATE =12

No of ways of arranging the consonants of which two are alike =$\large\frac{6!}{2!}$

$\Rightarrow \large\frac{6\times 5\times 4\times 3\times 2}{2}$

$\Rightarrow 360$

No of ways of arranging the vowels =$7P_6\times \large\frac{1}{3!}\times \frac{1}{2!}$

$\Rightarrow \large\frac{7!}{1!}\times \frac{1}{3!}\times \frac{1}{2!}$

$\Rightarrow \large\frac{7\times 6\times 5\times 4\times 3!}{3!\times 2}$

$\Rightarrow 420$

Total no of ways =$420\times 360$

$\Rightarrow 151200$

Hence (A) is the correct answer.

Ask Question

Tag:MathPhyChemBioOther

Take Test

...