Question

Let n and k be positive integer such that n ³$\frac{k(k + 1)}{2}$. The number of solution (x₁, x₂ …. x_k), x₁³ 1, x₂³ 2, ……. x_k³ k all integers satisfying x₁ + x₂ + … x_k = n is –

• A. ^{2n + k}C_n
• B. $2n + k + k^{2}C_{k - 1}$
• C. $2n + k + k^{2}C_{n}$
• D. None

Answer 1

Answer: (D)

None

Answer 2

No. of solution x₁ + x₂ + … x_k = n

= coefficient of xⁿ in (x + x² + x³ +… )

(x² + x³+ … ) …. (x^k + x^{k + 1} + ….)

= coefficient of xⁿ in x ^{1 + 2 + …. k} (1 + x + x² + …)^k

= coefficient of xⁿ in $x^{\frac{k(k + 1)}{2}}$ (1 – x)^–k

= coefficient of x^{n – r} in (1 – x)^–k $\frac{k(k + 1)}{2} = r$

coefficient of x^{n – r} in [1 + ^kC₁ x + ^{k + 1}C₂ x² + …]

^{k – 1 + n – r}C_{n – r} = ^{k – 1 + n – r}C_{k – 1} $\left\{ r = \frac{k(k + 1)}{2} \right\}$

k – 1 + n – r = k – 1 + n – r

k – 1 + n – $\frac{k(k + 1)}{2}$ = k – 1 + n – $\frac{k(k + 1)}{2}$

= $\frac{1}{2}$ (2n – k² + k – 2)

required no. of solⁿ = ^mC_{n – r} = ^mC_{k – 1}

where m = $\frac{1}{2}$ [2n – k² + k – 2]