Q.Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have

MAHARASHTRA has 11 letters: M,A,H,A,R,A,S,H,T,R,A — A appears 4 times, H appears twice, R appears twice, and M,S,T appear once each. Total arrangements $=\dfrac{11!}{4!\,2!\,2!}=\dfrac{39916800}{24\times2\times2}=\dfrac{39916800}{96}=415800$.

(a) 'R and H never together' means neither the two R's are adjacent to each other, NOR the two H's are adjacent to each other. Using inclusion–exclusion: let $A$ = arrangements where the two R's are together (bundle RR: $\dfrac{10!}{4!\,2!}=75600$), $B$ = arrangements where the two H's are together (bundle HH: $\dfrac{10!}{4!\,2!}=75600$), and $A\cap B$ = both pairs bundled ($\dfrac{9!}{4!}=15120$). Then $|A\cup B|=75600+75600-15120=136080$, and the arrangements with neither pair together are $415800-136080=279720$. …

For the total, divide by the factorial of every repeated letter's count. For 'never together' with two separate pairs, use inclusion–exclusion on the two 'together' events rather than treating them independently. For 'all togeth …

For part (a), computing only $415800-75600-75600$ (double-subtracting the overlap where both pairs are together) instead of adding back …