關於排列組合嘅幾條問題

七 (1 + 2x)ⁿ= 1ⁿ + (nC1)(2x) + (nC2)(2x)² + (nC3)(2x)³ + ... + (2x)ⁿ令 x = 1 , 各項係數和 = (1 + 2)ⁿ = 3ⁿ
八( x√x + 1/x⁴)ⁿ第三項的二項式係數 - 第二項的二項式係數 = 44
nC2 - nC1 = 44
n(n - 1)/2 - n = 44
n² - 3n - 88 = 0
(n - 11)(n + 8) = 0
n = 11 或 n = - 8(捨)( x√x + 1/x⁴)¹¹通項 T(n) = T(n-1 + 1)
= ( 11C(n-1) ) (x√x)¹¹⁻⁽ⁿ⁻¹⁾ (1/x⁴)ⁿ⁻¹
= ( 11C(n-1) ) (x¹⁺½)¹²⁻ⁿ (x⁻⁴)ⁿ⁻¹
= ( 11C(n-1) ) x^(18 - 1.5n + 4 - 4n)令 18 - 1.5n + 4 - 4n = 0
n = 4故不含 x 的常數項 = 11C(4-1) = 165
九 (1+x)³ + (1+x)⁴+ ... + (1+x)ⁿ⁺¹= (1+x)³ [ (1+x)ⁿ⁺¹⁻² - 1 ] / [ (1+x) - 1 ]= [ (1+x)ⁿ⁺² - (1+x)³ ] / x(1+x)ⁿ⁺² 展式中含 x³ 項的系數 = (n+2)C3
(1+x)³ 展式中含 x³ 項的系數 = 1故原展式含 x² 項的系數 = (n+2)C3 - 1
十(一)(nC0)² + (nC1)² + ... + (nCn)² = (2n)! / (n! n!)證 :考慮
nC0 + (nC1)x + ... + (nCn)xⁿ = (1 + x)ⁿ
及
(nC0)xⁿ + (nC1)xⁿ⁻¹ + ... + nCn = (x + 1)ⁿ 兩式相乘有 [ nC0 + (nC1)x + ... + (nCn)xⁿ ] [ (nC0)xⁿ + (nC1)xⁿ⁻¹ + ... + nCn ] = (1 + x)²ⁿ 比較兩方 xⁿ 的系數即得(nC0)² + (nC1)² + ... + (nCn)²
= 2n C n
= (2n)! / [(2n - n)! n!]
= (2n!) / (n! n!)
十(二) nC1 + 2(nC2) + 3(nC3) + ... + n(nCn) = n * 2ⁿ⁻¹證 :補空項 , 左方 =0(nC0) + nC1 + 2(nC2) + 3(nC3) + ... + n(nCn) 倒序相加 :2 左方 =

0(nC0) + nC1 + 2(nC2) + 3(nC3) + ... + n(nCn)
+
n(nCn) + (n-1)(nC(n-1) + (n-2)(nC(n-2) + (n-3)(nC(n-3)) + ... + 0(nC0)
利用恆等式 nCm = nC(n-m) ,
有 nC0 = nCn , nC1 = nC(n-1) , nC2 = nC(n-2) , nC3 = nC(n-3) , ....∴2 左方 =0(nC0) + nC1 +2(nC2) + 3(nC3) + ... + n(nCn)
+
n(nC0) + (n-1)(nC1) + (n-2)(nC2) + (n-3)(nC3) + ... + 0(nCn)= n (nC0 + nC1 + nC2 + nC3 + ... + nCn)= n (1 + 1)ⁿ= n * 2ⁿ故左方 = n * 2ⁿ / 2 = n * 2ⁿ⁻¹ = 右方