Exams › JEE Main › Maths

Evaluate the alternating sum (30 choose 0)(30 choose 10) - (30 choose 1)(30 choose 9) + (30 choose 2)(30 choose 8) -... + (30 choose 20)(30 choose 30), where (n choose r) denotes nCr.

1. (30 choose 10)
2. (30 choose 15)
3. (60 choose 30)
4. (31 choose 10)

Correct answer: (30 choose 10)

Solution

The alternating sum equals the coefficient of x^20 in (1+x)^30(1-x)^30 = (1-x^2)^30, which is (-1)^10 C(30,10) = C(30,10).

Related JEE Main Maths questions

• Let Pn = cosⁿ θ + sinⁿ θ. If Pn - Pn-2 = k Pn-4, then the value of k is:
• For any natural number n, the expression x^(2n - 1) + y^(2n - 1) is divisible by which of the following?
• For a natural number n, which of the following always divides the expression x^(n+1) + (x + 1)^(2n−1)?
• For a positive integer n, the expression 5^(2n+2) − 24n − 25 is divisible by which of the following?
• For every natural number n, the expression 41ⁿ − 14ⁿ is divisible by which of the following?
• Let C0, C1, C2,..., C15 denote the binomial coefficients in the expansion of (1 + x)¹⁵. Then the value of C1/C0 + C2/C1 + C3/C2 +... + C15/C14 is:

⚔️ Practice JEE Main Maths free + battle 1v1 →