Q.1. In the word MADE EASY, we have 6 distinct letters M, A, D, E, S, Y out of which A and E occur twice. Rest all occur once.
To select 4 letters, we have the following cases :
Case1: All 4 letters are distinct.
No. of ways = 6C4 = 6C2 = 15
Case 2: 2 same and 2 distinct
No. of ways = 2C1 * 5C2 = 20
Case 3: 2 groups of 2 same letters
No. of ways = 1 i.e. (A, A, E, E)
So, total no. of ways = 15 + 20 + 1 = 36
(a) is the answer
Q. 2. Now for no. of arrangements, we'll have :
Case 1 : All 4 distinct
No. of ways = 6C4 * 4! = 360
Case 2 : 2 same and 2 distinct
No. of ways = 4C2 (no. of positions the same letters will take) * 5C2 (ways of selecting distinct letters) * 2! (for arranging them) * 2 (because we can either select AA or EE as same letters) = 6 * 10 * 2 * 2 = 240
Case 3: 2 groups of 2 same letters
No. of ways = $\frac{4!}{2! 2!}$ = 6
$\therefore$ Total no. of ways = 360 + 240 + 6 = 606
(c) is the answer.
