Question

If the number of arrangements of letters of the word $DHARAMSHALA$ taken all at a time so that no two alike letters appear together is $4^a\cdot 5^b\cdot 6^c\cdot 7^d,$ where $a,b,c,d\in N,$ then the value of $a+b+c+d$ is

Answer 1

Given word $DHARAMSHALA$
$D-1,H-2,A-4,R-1,M-1,S-1,L-1$
Required number of ways
$=$ total number of ways when all $A^{\prime }s$ are seperated
$-$ total number of ways when all $A^{\prime }s$ are seperated and $H^{\prime }s$ are together
$=\frac{7!}{2!}\times {}^8{C}_4-6!\times {}^7{C}_4=6![\frac{7\times 8!}{2!4!4!}-\frac{7!}{4!3!}]=5\times 6\times 7![\frac{7\times 8}{2!4!}-\frac{1}{3!}]=5\times 6\times 7![\frac{56}{48}-\frac{1}{6}]=5\cdot 6\cdot 3!\cdot 4\cdot 5\cdot 6\cdot 7=4\cdot 5^2\cdot 6^3\cdot 7\Rightarrow a=1,b=2,c=3,d=1\therefore a+b+c+d=7$