Question

All letters of the word "BREAKAGE" are to be arranged. The number of ways of arranging them so that

Column I		Column II
A	the two A's are not together	P	1800
B	the two E's are together but not two A's	Q	5760
C	neither two A's nor two E's are together	R	6000
		S	7560

• A. (A) P ; (B) P, (C) Q
• B. (A) Q ; (B) P, (C) Q
• C. (A) R ; (B) P, (C) Q
• D. (A) S ; (B) P, (C) Q

Answer 1

Answer: (D)

$A =2 ; E =2 ; B =1 ; R =1 ; K =1 ; G =1$; total letters $=08$
(i) Number of words in which two A's are not together $=\left(\frac{8 !}{2 ! 2 !}-\frac{7 !}{2 !}\right)=7 !\left(2-\frac{1}{2}\right)=\frac{5040 \times 3}{2}=7560$ ]
(ii) Number of words in which two E's are together but not two A's
$
\left.=\left(\frac{7 !}{2 !}-6 !\right)=6 !\left(\frac{7}{2}-1\right)=\frac{720 \times 5}{2}=1800\right]
$
(iii) Number of words in which neither two A's nor two E's are together $=\left(\frac{8 !}{2 ! 2 !}-n(A \cup B)\right)=5760$ where $A \Rightarrow$ When $2 A$ 's are together and $ B \Rightarrow$ When 2 E's are together
$\therefore n(A \cup B)=n(A)+n(B)-n(A \cap B)=\left(\frac{7 !}{2 !}+\frac{7 !}{2 !}-6 !\right)$.