Hey guys can you check to see if my answer is correct? Thanks!!

Question:

Use inclusion-exclusion to find the number of solutions in the non-negative integers to

$\displaystyle X_1 + X_2 + $ .... $\displaystyle + X_6 = 45 $ with the conditions $\displaystyle X_j < 6$, $\displaystyle j = 1,2,3,4.$

My Answer:

Step 1: Find solutions with no conditions first.

$\displaystyle \binom{n+k-1}{k}$ $\displaystyle = \binom{6+45-1}{45}$ $\displaystyle = \binom{50}{45}$

Step 2: Get solutions for less than conditions

Condition 1: $\displaystyle X_j < 6 \Rightarrow X_j \leq 5 \Rightarrow X_j + 5$

Let:

$\displaystyle X_j = V_j + 5$

So: $\displaystyle (V_1 + 5) + (V_2 + 5) + (V_3 + 5) + (V_4 + 5) + X_5 + X_6 = 45$

$\displaystyle V_1 + V_2 + V_3 + V_4 + X_5 + X_6 = 25$

Therefore Solutions:$\displaystyle \binom{6+25-1}{25} = \binom{30}{25}$

THEREFORE FINAL SOLUTION $\displaystyle = \binom{50}{45} - \binom{30}{25}$

Is this correct? thanks