Question

Match each item from List-I with the correct answer from List-II:

• A. Number of ways in which 3 squares of unit length can be chosen on a 8 x 8 chessboard, so that all squares are in same diagonal line is
• B. Number of words that can be formed using the letters of the word 'LIVELIHOOD' if each word contains three letters is
• 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
• D. Number of ways in which the letters of the word 'ABBCABBC' can be arranged such that the word ABBC does not appear is any word is

Answer 1

(P) → 392 [Option (3)], (Q) → 264 [Option (4)], (R) → 441 [Option (1)], (S) → 361 [Option (2)]

Answer 2

We match each item from List‑I with the correct answer from List‑II as follows:

(P) Number of ways in which 3 squares on an 8×8 chessboard can be chosen so that they all lie on the same diagonal

A chessboard has two families of diagonals (in both the “/” and “\” directions). For one family (say “\” direction) the diagonals have lengths: 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1.

Only those diagonals with length L ≥ 3 contribute and the number chosen from a diagonal of length L is C(L, 3). For one direction the sum is: C(3,3)+C(4,3)+C(5,3)+C(6,3)+C(7,3)+C(8,3)+C(7,3)+C(6,3)+C(5,3)+C(4,3)+C(3,3) = 1+4+10+20+35+56+35+20+10+4+1 = 196. Since both diagonal directions are identical, total = 2×196 = 392.

Thus, (P) = 392 → Option (3).

(Q) Number of three‐letter words formed using the letters of “LIVELIHOOD”

Letters in LIVELIHOOD (10 letters): ‑ Frequency: L(2), I(2), O(2), and V, E, H, D (each 1).

Two cases:

1. All three letters distinct: There are 7 distinct letters. Arrangements = P(7,3) = 7×6×5 = 210.

2. Two letters identical and one different: Only letters with at least 2 copies (L, I, O) can be repeated. For any such letter, choose one different letter (from the remaining 6 distinct letters) and arrange A, A, B in 3!/2! = 3 ways. Total = 3×6×3 = 54.

Thus, total words = 210 + 54 = 264. So, (Q) = 264 → Option (4).

(R) Number of 5‑element subsets from the first 11 natural numbers with at least two consecutive numbers

Total 5‑element subsets = C(11, 5) = 462.

Subsets with no two consecutive numbers = C(11 – 5 + 1, 5) = C(7, 5) = 21.

So, subsets with at least one pair consecutive = 462 – 21 = 441. Thus, (R) = 441 → Option (1).

(S) Number of arrangements of “ABBCABBC” in which “ABBC” does NOT appear as a block

Total arrangements of “ABBCABBC” (8 letters with A(2), B(4), C(2)) are: = 8!/(2!4!2!) = 420.

Now count arrangements that DO contain “ABBC” as a contiguous block. Treat “ABBC” as a single entity (block X). Removing the letters in one block from the multiset: Original: A(2), B(4), C(2). After removing “A, B, B, C”: remaining: A(1), B(2), C(1) (4 letters). Now we arrange the 5 objects: X, A, B, B, C. Number of arrangements = 5!/(2!) = 60. However, note that if the arrangement contains two occurrences of “ABBC” (which happens only in the unique case “ABBCABBC”), it is over‐counted. Since the arrangement “ABBCABBC” appears in 2 ways in this scheme, subtract 1. Thus, arrangements with “ABBC” appearing = 60 – 1 = 59.

So, arrangements avoiding “ABBC” = 420 – 59 = 361. Thus, (S) = 361 → Option (2).

Summary: For (P), sum C(L,3) over diagonals of length ≥ 3 in both directions. For (Q), add arrangements with 3 distinct letters and with one pair of identical letters. For (R), subtract the no‐consecutive subsets from total subsets. For (S), compute total arrangements and subtract those having “ABBC” (using block method with an adjustment for double counting).