Question

The value of the expression \( {}^{47} C_4 + \sum_{j=1}^{5} {}^{52-j} C_3 \) is:

• A. \( {}^{52} C_3 \)
• B. \( {}^{51} C_4 \)
• C. \( {}^{52} C_4 \)
• D. \( {}^{51} C_3 \)

Hint:

Use the hockey-stick identity \( \sum_{m=k}^{n} {}^m C_k = {}^{n+1} C_{k+1} - {}^k C_{k+1} \) to simplify sums of combinations with consecutive indices.

Answer 1

The Correct Option is C

Step 1: Write down the given expression.
We need to evaluate: \[ {}^{47} C_4 + \sum_{j=1}^{5} {}^{52-j} C_3. \] This is a sum of combinations: \[ {}^{47} C_4 + {}^{51} C_3 + {}^{50} C_3 + {}^{49} C_3 + {}^{48} C_3 + {}^{47} C_3. \]
Step 2: Use the hockey-stick (Christmas stocking) identity.
The hockey-stick identity states that for integers \( r \geq 0 \), \( k \geq 0 \): \[ \sum_{m=k}^{n} {}^m C_k = {}^{n+1} C_{k+1}. \] Rewrite the sum \( \sum_{j=1}^{5} {}^{52-j} C_3 \):
For \( j = 1 \), \( 52 - j = 51 \),
For \( j = 5 \), \( 52 - j = 47 \).
So the sum is: \[ \sum_{j=1}^{5} {}^{52-j} C_3 = {}^{51} C_3 + {}^{50} C_3 + {}^{49} C_3 + {}^{48} C_3 + {}^{47} C_3. \] This is \( \sum_{m=47}^{51} {}^m C_3 \). Using the hockey-stick identity with \( k = 3 \), \( m \) from 47 to 51: \[ \sum_{m=47}^{51} {}^m C_3 = {}^{51+1} C_{3+1} - {}^{47} C_{3+1} = {}^{52} C_4 - {}^{47} C_4. \]
Step 3: Combine with the first term.
The original expression is: \[ {}^{47} C_4 + \sum_{j=1}^{5} {}^{52-j} C_3 = {}^{47} C_4 + ({}^{52} C_4 - {}^{47} C_4) = {}^{52} C_4. \]
Step 4: Verify with a numerical check.
For \( n = 47 \), compute directly:
\( {}^{47} C_4 = \frac{47 \times 46 \times 45 \times 44}{4 \times 3 \times 2 \times 1} = 178365 \),
Sum: \( {}^{51} C_3 = 20825 \), \( {}^{50} C_3 = 19600 \), \( {}^{49} C_3 = 18424 \), \( {}^{48} C_3 = 17296 \), \( {}^{47} C_3 = 16215 \),
Total sum = \( 20825 + 19600 + 18424 + 17296 + 16215 = 92360 \),
Expression = \( 178365 + 92360 = 270725 \),
\( {}^{52} C_4 = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725 \), which matches.

Step 5: Select the correct answer.
The expression simplifies to \( {}^{52} C_4 \), matching option (C).