The number of permutations of all the letters | vedclass

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(A) Total letters are $8$ ($4$ $A$'s,$3$ $B$'s,$1$ $C$).Let $P$ be the set of permutations where all $4$ $A$'s appear together. Treating $AAAA$ as one

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(A) Total letters are $8$ ($4$ $A$'s,$3$ $B$'s,$1$ $C$).Let $P$ be the set of permutations where all $4$ $A$'s appear together. Treating $AAAA$ as one unit,we have $5$ units $(AAAA, B, B, B, C)$. The number of arrangements is $n(P) = \frac{5!}{3!} = \frac{120}{6} = 20$.Let $Q$ be the set of permutations where all $3$ $B$'s appear together. Treating $BBB$ as one unit,we have $6$ units $(A, A, A, A, BBB, C)$. The number of arrangements is $n(Q) = \frac{6!}{4!} = \frac{720}{24} = 30$.Let $P \cap Q$ be the set of permutations where both $AAAA$ and $BBB$ appear together. Treating $AAAA$ as one unit and $BBB$ as one unit,we have $3$ units $(AAAA, BBB, C)$. The number of arrangements is $n(P \cap Q) = 3! = 6$.Using the Principle of Inclusion-Exclusion,$n(P \cup Q) = n(P) + n(Q) - n(P \cap Q) = 20 + 30 - 6 = 44$.

The number of permutations of all the letters $AAAABBBC$ in which all the $A$'s appear together in a block of $4$ letters or all the $B$'s appear together in a block of $3$ letters,is-

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