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 $=2\left({ }^8 C _3\right)+4\left({ }^7 C _3+{ }^6 C _3+{ }^5 C _3+{ }^4 C _3+{ }^3 C _3\right)$ $=2 \times
56+4(35+20+10+4+1)=112+280=392$ 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 $\times A B B C$ and $A B B C \times$.
But the word ABBCABBC is counted twice in 60 hence it should be 59
Hence required number of ways $=420-59=361$