Question

If $\lambda$ is the number of $7$ letter words formed by using all the letters of the word 'AGAIN' such that all three vowels are taken in such a way that no two vowels are together and only consonants can be repeated as many times as we need, then $\frac{\lambda }{210}$ is

Answer 1

Here $3$ vowels $A,I,A$ can be placed in the following five available places.
$\times N\times G\times N\times N\times$
OR
$\times N\times G\times G\times G\times$
OR
$\times N\times G\times G\times N\times$
$\therefore$ Required number of ways,
$\lambda ={}^5{C}_3\times \frac{3!}{2!}\times [\frac{4!}{3!}+\frac{4!}{3!}+\frac{4!}{2!\times 3!}]$
$=\frac{5\times 4\times 3!}{3!\times 2\times 1}\times \frac{3\times 2!}{2!}[\frac{4\times 3!}{3!}+\frac{4\times 3!}{3!}\times \frac{4\times 3\times 2!}{2!\times 2}]$
$=30(4+4+6)=420$
$\therefore \frac{\lambda }{210}=\frac{420}{210}=2$