If ²ⁿ C₃: ⁿ C₃=10: 1, then the ratio (n²+3 n):(n²-3 n+4) is :

Solution

Maths sir · 5 steps

Write both as products, cancel the common factors, and n drops out of a linear equation.

1. Expand both. Use ʳC₃ = r(r-1)(r-2)/6. The 6 cancels in the ratio, so only the products of three consecutive numbers remain.

   ²ⁿC₃/ⁿC₃ = 2n(2n-1)(2n-2)/n(n-1)(n-2) = 4(2n-1)/(n-2)

2. Cancel carefully. Write 2n-2 = 2(n-1). Then 2n cancels with n to leave 2, and 2(n-1) cancels with (n-1) to leave another 2. That gives the factor 4.
3. Solve for n. Set 4(2n-1)/(n-2) = 10. Then 8n - 4 = 10n - 20, so 2n = 16 and n = 8.
4. Substitute. With n = 8: n² + 3n = 64 + 24 = 88 and n² - 3n + 4 = 64 - 24 + 4 = 44.
5. So. The ratio is 88 : 44 = 2 : 1, which is option (b).

The idea to keep. For a ratio of combinations with the same r, the r! cancels. Factor out common numbers like 2n-2 = 2(n-1) before cancelling, and the equation in n turns linear.

Answer from the NTA final answer key. Solution by PrepWiser.