Question

If for \( 3 \leq r \leq 30 \), \( ^{30}C_{30-r} + 3 \left( ^{30}C_{31-r} \right) + 3 \left( ^{30}C_{32-r} \right) + ^{30}C_{33-r} = ^m C_r \), then \( m \) equals to_________ 

Hint:

Whenever you see coefficients like 1, 2, 1 or 1, 3, 3, 1 in a combination sum, it indicates that Pascal's identity is being applied 2 or 3 times respectively. The upper index \( n \) will increase by the power of the expansion.

Answer 1

Correct Answer: 33

Step 1: Understanding the Concept:
This problem utilizes Pascal's Identity for binomial coefficients, which states that \( ^nC_r + ^nC_{r-1} = ^{n+1}C_r \). The given expression is a series of combinations that can be simplified by repeatedly applying this identity.
Step 2: Key Formula or Approach:
The coefficients (1, 3, 3, 1) in the expression suggest a binomial expansion of \( (1+x)^3 \). We can group the terms to apply Pascal's Rule: \[ ^nC_r + ^nC_{r-1} = ^{n+1}C_r \] We also use the property \( ^nC_r = ^nC_{n-r} \) to simplify indices if necessary.
Step 3: Detailed Explanation:
First, let \( n = 30 \). Using \( ^nC_k = ^nC_{n-k} \), the terms are: \[ ^{30}C_r + 3(^{30}C_{r-1}) + 3(^{30}C_{r-2}) + ^{30}C_{r-3} \] Group them as follows: \[ \left[ ^{30}C_r + ^{30}C_{r-1} \right] + 2 \left[ ^{30}C_{r-1} + ^{30}C_{r-2} \right] + \left[ ^{30}C_{r-2} + ^{30}C_{r-3} \right] \] Applying Pascal's Rule to each bracket: \[ ^{31}C_r + 2(^{31}C_{r-1}) + ^{31}C_{r-2} \] Regroup again: \[ \left[ ^{31}C_r + ^{31}C_{r-1} \right] + \left[ ^{31}C_{r-1} + ^{31}C_{r-2} \right] \] Applying Pascal's Rule again: \[ ^{32}C_r + ^{32}C_{r-1} \] Applying Pascal's Rule one last time: \[ ^{33}C_r \] Comparing \( ^{33}C_r \) with \( ^mC_r \), we find \( m = 33 \).
Step 4: Final Answer:
The value of \( m \) is 33.