The problem:
We need to distribute ten identical chocolates among five children - V, W, X, Y, and Z. The distribution should be such that only two children get exactly two chocolates each, while the rest can get any number of chocolates.

Approach:
To solve this problem, we can break it down into different cases based on the number of chocolates received by the remaining three children - Y, Z, and W. Let's consider each case separately:

Case 1: Y, Z, and W receive 0 chocolates each.
In this case, V and X are the only two children who receive two chocolates each. The remaining six chocolates can be distributed in various ways among V, X, Y, Z, and W.

- The first chocolate can be given to any of the five children.
- The second chocolate can be given to any of the remaining four children.
- The third chocolate can be given to any of the remaining three children.
- The fourth chocolate can be given to any of the remaining two children.
- The fifth chocolate can be given to any of the remaining one child.
- The sixth chocolate will be automatically given to the last remaining child.

Therefore, the number of ways to distribute six chocolates among five children (assuming Y, Z, and W receive 0 chocolates each) is 5 * 4 * 3 * 2 * 1 = 120.

Case 2: Y receives 1 chocolate, while Z and W receive 0 chocolates each.
In this case, V and X receive two chocolates each. Y receives one chocolate, and Z and W receive 0 chocolates each. The remaining five chocolates can be distributed among V, X, Y, Z, and W.

- The first chocolate can be given to any of the five children.
- The second chocolate can be given to any of the remaining four children.
- The third chocolate can be given to any of the remaining three children.
- The fourth chocolate can be given to any of the remaining two children.
- The fifth chocolate will be automatically given to the last remaining child.

Therefore, the number of ways to distribute five chocolates among five children (assuming Y receives 1 chocolate, and Z and W receive 0 chocolates each) is 5 * 4 * 3 * 2 = 120.

Case 3: Y and Z receive 1 chocolate each, while W receives 0 chocolates.
In this case, V and X receive two chocolates each. Y and Z receive one chocolate each, and W receives 0 chocolates. The remaining four chocolates can be distributed among V, X, Y, Z, and W.

- The first chocolate can be given to any of the five children.
- The second chocolate can be given to any of the remaining four children.
- The third chocolate can be given to any of the remaining three children.
- The fourth chocolate will be automatically given to the last remaining child.

Therefore, the number of ways to distribute four chocolates among five children (assuming Y and Z receive 1 chocolate each, and W receives 0 chocolates) is 5 * 4 * 3 = 60.

Case 4: Y, Z, and W receive 1 chocolate each.
In this case, V and X receive two chocolates each. Y, Z, and W receive one chocolate each. The