Properties of Binomial Coefficients

[MathDefinition title="Binomial Coefficients"] Binomial coefficients nCr are the coefficients in the expansion of (a + b)^n. They appear in Pascal's triangle and have many important properties. [/MathDefinition]

[MathKeyFormula title="Key Properties"]

1. nC0 = nCn = 1
2. nC1 = nCn-1 = n
3. nCr = nCn-r (Symmetry property)
4. nCr + nCr-1 = n+1Cr (Pascal's identity)
5. Sum of coefficients: nC0 + nC1 + ... + nCn = 2^n
6. Alternating sum: nC0 - nC1 + nC2 - ... = 0
7. nC0 + nC2 + nC4 + ... = nC1 + nC3 + nC5 + ... = 2^(n-1)
8. n * nCr = (n-r+1) * nCr-1 [/MathKeyFormula]

[MathNote title="Using Properties"] These properties simplify calculations and help prove identities. The symmetry property is especially useful for reducing computation. [/MathNote]

[MathExample title="Example 1: Sum of Coefficients"] Problem: Find the sum of all binomial coefficients in the expansion of (1 + x)^10.

Solution: The sum of all binomial coefficients is: nC0 + nC1 + nC2 + ... + nCn = 2^n

Here n = 10 Sum = 2^10 = 1024

Therefore, the sum of all binomial coefficients is 1024. [/MathExample]

[MathExample title="Example 2: Using Pascal's Identity"] Problem: If 15Cr = 15Cr-3, find r.

Solution: Using the property nCr = nCn-r, we have: Either r = n - r + 3 or r = r - 3 (impossible)

So r = n - r + 3 2r = n + 3 2r = 15 + 3 = 18 r = 9

Alternatively, using the property directly: 15Cr = 15Cr-3 implies either: r + (r-3) = 15 (i.e., r + n - r = n), giving 2r - 3 = 15, r = 9 Or r = r - 3, which is impossible.

Therefore, r = 9. [/MathExample]

[MathExample title="Example 3: Proving Identity"] Problem: Prove that nC0 + nC2 + nC4 + ... = 2^(n-1).

Solution: We know that (1 + 1)^n = nC0 + nC1 + nC2 + ... + nCn = 2^n

And (1 - 1)^n = nC0 - nC1 + nC2 - nC3 + ... = 0

Adding these two equations: 2^n + 0 = 2(nC0 + nC2 + nC4 + ...)

Therefore, nC0 + nC2 + nC4 + ... = 2^n / 2 = 2^(n-1)

Hence proved. [/MathExample]

[MathExample title="Example 4: Finding Ratio"] Problem: If nC3 : nC5 = 5 : 3, find n.

Solution: Given nC3 : nC5 = 5 : 3

nC3 = n!/(3!(n-3)!) nC5 = n!/(5!(n-5)!)

nC3/nC5 = 5!/3! * (n-5)!/(n-3)! = 20/[(n-3)(n-4)]

Given ratio = 5/3: 20/[(n-3)(n-4)] = 5/3

60 = 5(n-3)(n-4) 12 = (n-3)(n-4) n^2 - 7n + 12 = 12 n^2 - 7n = 0 n(n-7) = 0

Since n cannot be 0, n = 7. [/MathExample]