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
$<br/>=\left(\frac{7!}{2!}-6!\right)=6!\left(\frac{7}{2}-1\right)=\frac{720\times 5}{2}=1800]<br/>$
(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 $2A$ 's are together and $B\Rightarrow$ When 2 E's are together
$\therefore n(A\cup B)=n(A)+n(B)-n(A\cap B)=\left(\frac{7!}{2!}+\frac{7!}{2!}-6!\right)$.