If the coefficient of a⁷b⁸ in the expansion of (a + 2b + 4ab)¹⁰ is K.2¹⁶, then K is equal to _____________.

Solution

Maths sir · 4 steps

Write a general term, then match the powers of a and b to fix how many times each bracket term is chosen.

1. General term. Choose a exactly p times, 2b exactly q times and 4ab exactly r times, with p + q + r = 10. That term is 10!/(p! q! r!) aᵖ (2b)^q (4ab)ʳ, which contains aᵖ⁺ʳ b^(q+r).
2. Match the powers. We need p + r = 7 and q + r = 8. Adding these gives p + q + 2r = 15. Since p + q + r = 10, this means r = 5, so p = 2 and q = 3.

   10!/(2! 3! 5!) × 2³ × 4⁵ = 2520 × 2³ × 2¹⁰ = 2520 × 2¹³

3. Match with K. Set 2520 × 2¹³ = K × 2¹⁶. Then K = 2520/2³ = 2520/8 = 315.
4. So. The coefficient of a⁷ b⁸ is 315 × 2¹⁶, so K = 315.

The common slip: 2520. Taking the multinomial coefficient as K gives 2520. The powers of 2 total only 2¹³, so you must divide by 2³ to match 2¹⁶.

The idea to keep. In a multinomial, a term like ab adds to both powers at once. Write one equation for each variable's power and one for the total count, then solve for how often each term is picked.

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