Question

Let tk be the number of ways to choose k beads from a drawer that contains 10 red beads, 8 blue beads, and 11 green beads. Find the generating function for the sequence {tk} 29 k = 0. Do not find a formula for tk.

Answer 1

Answer:

$\mathbf{G(x) = (1 + x)^{29}}$

Step-by-step explanation:

From the given information;

The total number of beads  = 10 + 8 + 11 = 29

To choose k beads out 29 beads, let assume that $t_k$ be the no. such that:

$t_k = ( ^{29}_{k}) \ \ \ where; k =0,1,2,3,....,29$

∴

The sequence $\{ t_k\}^{29}_{k=0} = (^{29}_{k})$

The generating function for $\{ a_k \}^n_{k=0}$ is:

$G(x) = a_o +a_1x +a_2x^2 +...+a_nx^n$

$G(x) = \sum \limits ^{n}_{k=0} \ a_k x^k$

For the sequence above;

$G(x) = \sum \limits ^{29}_{k =0 } (^{29}_{k}) x^k$

By applying binomial series $(1+x)^n = \sum \limits ^n_{k=0} ( ^n_k) x^k$

Thus:

$\mathbf{G(x) = (1 + x)^{29}}$