If nCr : nCr+1 = 1 : 2  and nCr+1 : nCr+2 = 2 : 3  then find the value of n and r?

Concept:

Properties of Binomial Coefficients:

• Binomial coefficients are written as ⁿC_r and represent combinations.
• Formula: ⁿC_r = n! / [r!(n − r)!]
• Use ratio of consecutive terms to simplify:
• \( \frac{{^nC_{r+1}}}{{^nC_r}} = \frac{n - r}{r + 1} \)  
• \(\frac{{^nC_{r+2}}}{{^nC_{r+1}}} = \frac{n - r - 1}{r + 2} \) 

Calculation:

Given,

ⁿC_r : ⁿC_r+1 = 1 : 2

⇒ ⁿC_r+1 / ⁿC_r = 2

⇒ $\frac{n - r}{r + 1} = 2$

⇒ n − r = 2r + 2

⇒ n = 3r + 2  ......(i)

Also given,

ⁿC_r+1 : ⁿC_r+2 = 2 : 3

⇒ ⁿC_r+2 / ⁿC_r+1 = 3/2

⇒ $\frac{n - r - 1}{r + 2} = \frac{3}{2}$

Cross-multiplying:

⇒ 2(n − r − 1) = 3(r + 2)

⇒ 2n − 2r − 2 = 3r + 6

⇒ 2n = 5r + 8  ......(ii)

Substitute equation (i) in (ii):

2(3r + 2) = 5r + 8

⇒ 6r + 4 = 5r + 8

⇒ r = 4

Substitute r = 4 in equation (i):

n = 3×4 + 2 = 14

∴ The values are: n = 14, r = 4