Question

If \[ {}^{n-3}C_r+B\,{}^{n-3}C_{r-1}+B^1\,{}^{n-3}C_{r-2}+{}^{n-3}C_{r-3}={}^nC_r \] holds for all \(n\geq r\geq 3\), then \((B,B^1)=\)

• A. \((1,5)\)
• B. \((5,1)\)
• C. \((3,3)\)
• D. \((4,2)\)

Hint:

Use Pascal's identity repeatedly to express \({}^nC_r\) in terms of combinations with a smaller upper index.

Answer 1

The Correct Option is C

Step 1: Recall Pascal's identity.
ⁿC_r = ⁿ⁻¹C_r + ⁿ⁻¹C_{r-1}. This recurrence can be applied repeatedly to lower the upper index.

Step 2: Expand ⁿC_r down to index n-2.
First expansion: ⁿC_r = ⁿ⁻¹C_r + ⁿ⁻¹C_{r-1}. Expanding each term: ⁿ⁻¹C_r = ⁿ⁻²C_r + ⁿ⁻²C_{r-1} and ⁿ⁻¹C_{r-1} = ⁿ⁻²C_{r-1} + ⁿ⁻²C_{r-2}. Combining: ⁿC_r = ⁿ⁻²C_r + 2·ⁿ⁻²C_{r-1} + ⁿ⁻²C_{r-2}.

Step 3: Expand once more to reach index n-3.
Apply Pascal's identity to each term: ⁿ⁻²C_r = ⁿ⁻³C_r + ⁿ⁻³C_{r-1}, ⁿ⁻²C_{r-1} = ⁿ⁻³C_{r-1} + ⁿ⁻³C_{r-2}, and ⁿ⁻²C_{r-2} = ⁿ⁻³C_{r-2} + ⁿ⁻³C_{r-3}. Summing with coefficients yields ⁿC_r = ⁿ⁻³C_r + 3·ⁿ⁻³C_{r-1} + 3·ⁿ⁻³C_{r-2} + ⁿ⁻³C_{r-3}.

Step 4: Match coefficients with the given form.
Comparing with ⁿC_r = ⁿ⁻³C_r + B·ⁿ⁻³C_{r-1} + B¹·ⁿ⁻³C_{r-2} + ⁿ⁻³C_{r-3}, we identify B = 3 and B¹ = 3.

Step 5: Final conclusion.
The ordered pair is (B, B¹) = (3, 3).