In how many ways can `2t+1` identical balls be placed in three distinct boxes so that any two boxes together will contain more balls than the third?

Explanation

The total number of ways to place the balls disregarding the constrains is `""^(2t+1+3-1)C_(3-1)= ""^(2t+3)C_(2)`.
The total number of ways to place the balls so that the first box will have balls than the other two is
`""^(t+3-1)C_(3-1)=""^(t+2)C_(2)`.
[We place t+1 balls in the first box and then divide the rest of t balls in the three boxes arbitrarily.]