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:

[tex]\mathbf{G(x) = (1 + x)^{29}}[/tex]

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 [tex]t_k[/tex] be the no. such that:

[tex]t_k = ( ^{29}_{k}) \ \ \ where; k =0,1,2,3,....,29[/tex]

∴

The sequence [tex]\{ t_k\}^{29}_{k=0} = (^{29}_{k})[/tex]

The generating function for [tex]\{ a_k \}^n_{k=0}[/tex] is:

[tex]G(x) = a_o +a_1x +a_2x^2 +...+a_nx^n[/tex]

[tex]G(x) = \sum \limits ^{n}_{k=0} \ a_k x^k[/tex]

For the sequence above;

[tex]G(x) = \sum \limits ^{29}_{k =0 } (^{29}_{k}) x^k[/tex]

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

Thus:

[tex]\mathbf{G(x) = (1 + x)^{29}}[/tex]

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