`sum_(r=0)^(20).^(50-r)C_6`

Answer

To solve the problem \( \sum_{r=0}^{20} \binom{50-r}{6} \), we can follow these steps:

### Step 1: Rewrite the Summation
We start by rewriting the summation to make it easier to compute:
\[ \sum_{r=0}^{20} \binom{50-r}{6} \]
We can change the index of summation by letting \( k = 20 - r \). Thus, when \( r = 0 \), \( k = 20 \) and when \( r = 20 \), \( k = 0 \). This gives us:
\[ \sum_{k=0}^{20} \binom{50-(20-k)}{6} = \sum_{k=0}^{20} \binom{30+k}{6} \]

### Step 2: Change the Limits of Summation
Now we can rewrite the summation:
\[ \sum_{k=0}^{20} \binom{30+k}{6} \]

### Step 3: Use the Hockey Stick Identity
We can apply the hockey stick identity (or Christmas stocking theorem) which states:
\[ \sum_{j=r}^{n} \binom{j}{r} = \binom{n+1}{r+1} \]
In our case, we have:
\[ \sum_{k=0}^{20} \binom{30+k}{6} = \binom{30+20+1}{6+1} = \binom{51}{7} \]

### Step 4: Final Calculation
Now we have:
\[ \sum_{r=0}^{20} \binom{50-r}{6} = \binom{51}{7} \]

### Step 5: Subtract the Extra Terms
We need to account for the terms we originally excluded. The original summation was:
\[ \sum_{r=0}^{20} \binom{50-r}{6} = \binom{51}{7} - \binom{30}{7} \]

### Final Answer
Thus, the final answer is:
\[ \sum_{r=0}^{20} \binom{50-r}{6} = \binom{51}{7} - \binom{30}{7} \]

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