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Let n and k be positive such that n ≥ k (k + 1)/2. The number of solutions (x 1 , x 2 , . . . . . . . . . . . . . . .x k ) x 1 ≥ 1, x 2 ≥ 2, . . . . . . . . . . . . . . . ., x k ≥ k, all integers, satisfying x 1 + x 2 + . . . . . . . . + x k = n , is . . . . . . . .

Let n and k be positive such that n ≥ k (k + 1)/2. The  number of solutions (x1, x2,  . . . . . . . . . . . . . . .xk) x1 ≥ 1, x2 ≥ 2, . . . . . . . . . . . . . . . ., xk ≥ k, all integers, satisfying x1 + x2 + . . . . . . . . + xk = n , is . . . . . . . .

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
7 years ago
We have
X1 + x2 + . . . . . . . . . . . . . . . . . . . . . xk = n . . . . . . . . . . . (1)
Where x1 ≥ 1, x2 ≥ 2, x3 ≥ 3 . . . . . . . . . . . . . . ., xk ≥ k ; all integers
Let y1 = x1 – 1, y2 = x2 – 2, . . . . . . . . . . . . . . . yk = xk – k
So that y1, y2, . . . . . . . . . . . . . . , yk ≥ 0
Substituting the values of x1, x2, . . . . . . . . . . . . . . .xk in equation . . . . . . . . . . . . . . . . (1)
We get y1 + y2 + . . . . . . . . yk = n – (1 + 2 + 3 + . . . . . . k)
= n – k(k + 1)/2 . . . . . . . . . . . . . (2)
Now keeping in mind that number of solutions of the equation
α + 2β + 3γ + . . . . . . . . . . . qθ = n
for α, β, γ, . . . . . . . . . . . . . θ ϵ I and each is ≥ 0, given by coeff of xn in
(1 + x + x2 + . . . . . . . . . . . ) (1 + x2 + x4 + . . . . . . . . . . . )
(1 + x3 +x6 + . . . . . . . . . . . . . . .) . . . . . . . . . . . .( 1 + xq + x2q + . . . . . . . . . . . )
We find that no. of solutions of equation (2)
= coeff of xn – k(k + 1)/2 in (1 + x + x2 + . . . . . . . . . . . . . )k
NOTE THIS STEP
= coeff of xn – k(k + 1)/2 in (1 – x)-k
= coeff of xn – k(k + 1)/2 in ( 1+ kC1 x + k + 1C2 x2 + k + 2C3x3 + . . . . . . . . . . . . .) = k + (n – k(k +1)/2) – 1Cn – k(k + 1)/2
= [k + (n – k(k + 1)/2) – 1]!/[ n – k(k + 1)/2]! (k – 1)!

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