Four boys picked up 30 mangoes .In how many ways can they

1. If boys can get zero mangoes them the answer would be - \((30+4-1)C(4-1)=33C3\):

Consider 30 Mangoes: ****************************** and 3 separators |||.
Permutations of these 33 symbols out of which 30 *'s and 3 |'s are identical is \(\frac{33!}{3!30!}\), or written in another way \(C^3_{33}\).

Each permutation will mean one particular distribution of 30 mangoes among 4 boys:
******************************||| first boy gets all mangoes;
*|*|*|*************************** first, second and third boys get 1 mango each, and fourth gets 27;
*||*|*************************** first gets one mango, second boy gets zero, third boy get 1 and fourth gets 28;
And so on.

2. If boys should get at least one mango then the answer would be - \((26+4-1)C(4-1)=29C3\) (basically we are distributing 26 mangoes):

The same as above: we should jut give 1 mango to each boy and then distribute 26 mangoes left as in previous case.

26 Mangoes: ************************** and 3 separators |||.
Permutations of these 29 symbols out of which 26 *'s and 3 |'s are identical is \(\frac{29!}{3!26!}\), or written in another way \(C^3_{29}\).

Again each permutation will mean one particular distribution of 26 mangoes among 4 boys.

Similar problems:
voucher-98225.html?hilit=separators
integers-less-than-85291.html?hilit=identical#p710836

Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).

P.S. 4^30 would be the answer if all mangoes were different but we are told that they are identical.

Hope it helps.