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

Question

The number of permutations of all the letters AAAABBBC in which all the A's appear together in the block of 3 letters, is- (A) 44 (B) 50 (C) 60 (D) 89

Tardigrade Admin · Accepted Answer

P = All A' s together =(5 !/3 !), Q = all B 's together =(6 !/4 !) n ( P ∩ Q )=3 ! ⇒ n ( P ∪ Q ) =(5 !/3 !)+(6 !/4 !)-3 !=44

Q. The number of permutations of all the letters $AAAA BBBC$ in which all the $A^{'} s$ appear together in the block of $3$ letters, is-

44

50

60

89

$P =$ All $A^{'} s$ together $= \frac{5 !}{3 !}$ ,
$Q =$ all $B$ 's together $= \frac{6 !}{4 !}$
$n (P \cap Q) = 3! \Rightarrow n (P \cup Q)$
$= \frac{5 !}{3 !} + \frac{6 !}{4 !} - 3! = 44$

Q. The number of permutations of all the letters AAAABBBC in which all the A′s appear together in the block of 3 letters, is-

44

50

60

89

P= All A′s together =3!5!​, Q= all B 's together =4!6!​ n(P∩Q)=3!⇒n(P∪Q) =3!5!​+4!6!​−3!=44

Q. The number of permutations of all the letters $AAAA BBBC$ in which all the $A^{'} s$ appear together in the block of $3$ letters, is-

$P =$ All $A^{'} s$ together $= \frac{5 !}{3 !}$ ,
$Q =$ all $B$ 's together $= \frac{6 !}{4 !}$
$n (P \cap Q) = 3! \Rightarrow n (P \cup Q)$
$= \frac{5 !}{3 !} + \frac{6 !}{4 !} - 3! = 44$