Question

Column I		Column II
A	Number of ways in which 3 squares of unit length can be chosen on a $8\times 8$ chessboard, so that all squares are in same diagonal line, is	P	360
B	Number of words that can be formed using the letters of the word "LIVELIHOOD", if each word contains three letters, is	Q	361
C	Number of five element subset that can be chosen from the set of the first 11 natural numbers so that at least two of the five elements are consecutive, is	R	392
D	Number of ways in which the letters of the word 'ABBCABBC' can be arranged such that the word ABBC does not appear in any word, is	S	264
		T	441

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

Answer 1

Answer: (A)

(A) Total ways $=2\left({}^8{C}_3\right)+4\left({}^7{C}_3+{}^6{C}_3+{}^5{C}_3+{}^4{C}_3+{}^3{C}_3\right)$ $=2\times <br/>56+4(35+20+10+4+1)=112+280=392$ Ans.

(B) $L-2;I-2;O-2;H-1;D-1;V-1,E-1$

(C) $S=\{1,2,3,4,5,6,7,8,9,10,11\}$
Any five elements can be selected in ${}^{11}{C}_5=462$ ways
Now number of ways in which five elements from the set $S$ can be selected so that no two of them are consecutive is ${}^7{C}_5=21$
$|\times |\times |\times |\times |\times |\times \mid \text{ (7 gaps) }$
$\therefore$ Required number of ways $={}^{11}{C}_5-{}^7{C}_5=462-21=441$ Ans.
(D) $A$ 's $=2;B^{\prime }s=4;C^{\prime }s=2$
Total words formed $=\frac{8!}{4!2!2!}=420$
Let $ABBC=$ ' $x$ '
Number of ways in which $\times$ ABBC can be arranged $=\frac{5!}{2!}=60$ but this includes $\times ABBC$ and $ABBC\times$.
But the word ABBCABBC is counted twice in 60 hence it should be 59
Hence required number of ways $=420-59=361$