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

Single-Digit Nos. = 2; 

2-Digit Nos. = 2 (Ways of selecting unit’s digit) x 3 (Ways of selecting ten’s digit as zero cannot be used) = 6; 3-digit Nos. = 2 (Ways of selecting unit’s digit) x 4 (Ways of selecting ten’s digit) x 3 (Ways of selecting hundred’s digit as zero cannot be used) = 24;

4-digit Nos. = 2 (Ways of selecting unit’s digit) x 4 (Ways of selecting ten’s digit) x 4 (Ways of selecting hundred’s digit) x 3 (Ways of selecting thousand’s digit as zero cannot be used) = 96;

∴Total = 2 + 6 + 24 + 96 = 128