Total letters = 8; A ? 2, E ? 1, I ?1 , D ? 1, M ? 1, C ? 2. Required number of permutations = 8!/ (2!2!)

Alternate positions can be either all odd or all even, which means two ways. The rest of the letters can be arranged in the left over positions in 4! Ways. Hence, total number of arrangements possible = 2 x 4! = 48

The caps are identical and ladies are different. So the different arrangements get generated due to varying numbers of garlands worn by different ladies. 

If all the caps are worn by just one person out of four ⇒ No. of ways = 4 

If all the caps are worn by two ladies out of four; Two people can be selected in 4C2 = 6 ways. Now, these two people can wear 6 caps in the following ways (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) i.e. 5 ways. Therefore, number of ways = 5 x 6 = 30

If all the caps are worn by three ladies out of four; Three ladies can be selected in 4C3 = 4 ways. In each of the selections, we can have the following combinations: (1, 1, 4) i.e. 3!/2! = 3 ways; (2, 2, 2) i.e. 1 ways;   (3, 2, 1) i.e. 3! = 6 ways. Therefore, number of ways = 4 x (3 + 1 + 6) = 40 ways.

If all the caps  are worn by four ladies out of four; We can have the following arrangements: (1, 1, 2, 2) which can be done in 4!/(2!2!) = 6 ways and (3, 1, 1, 1) which can be done in 4!/3! = 4 ways.

Hence, the total number of arrangements possible = 4 + 30 + 40 + 10 = 84 ways.

Alternatively, no. of ways = n + r – 1C n – 1. Here, n = 4 and r = 6

Since the caps are different, we can consider it this way. Each cap can go to any of the four ladies. So, there are 4 ways of wearing one cap. Corresponding to each of this way, there are 4 ways of wearing the second caps. So the total number of ways of wearing two “different” caps = 4 x 4. Extending the argument for 6 caps, there are 46 ways of wearing 6 different caps by 4 ladies.

Required number of ways = Total number of ways – Number of ways in which he does not pick any defective piece = 15C3 – 8C3 = (15 x 14 x 13 –  8 x 7 x 6 )/6 = 399