CAT 2007 — QA Question 4

Forming a committeeEasy

Passage / Data

Answer the next 2 questions based on the information given below.

Let S be the set of all pairs (i, j) where 1 ≤ i < j ≤ n and n ≥ 4. Any two distinct members of S are called “friends” if they have one constituent of the pairs in common and “enemies” otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1, 2) and (2, 3) are also friends, but (1, 4) and (2, 3) are enemies.

For general n, how many enemies will each member of S have?

Answer & solution

A
n – 3
B
$\frac{1}{2} (n^{2} - 3 n - 2)$
C
2n - 7
$\frac{1}{2} (n^{2} - 5 n + 6)$
E
$\frac{1}{2} (n^{2} - 7 n + 14)$

Solution

Enemies of every pair are the pairs formed with all numbers other than the two in the member itself.

∴ If there are n elements then each member has

$^{(n - 2)} C_{2} = \frac{(n - 2) (n - 3) (n - 4)!}{2 (n - 4)!} = \frac{1}{2} (n^{2} - 5 n + 6) enemies$

Hence, option (d).

CAT 2007 QA Q4: For general n, how many enemies will each member of S have? — Solution | TheCATExam