(a) Choose `3` from `9`, since the eldest boy is fixed:

`C_3^9 ` `= frac{9!}{3!(9-3)!}` `=frac{9!}{3! times 6!}` `= 84`

(b) If the eldest boy is excluded, it is actually choose `4` boys from `9`:

`C_4^9 ` `= frac{9!}{4!(9-4)!}` `=frac{9!}{4! times 5!}` `= 126`

(c) The number of all possible groups is

`C_4^10 = frac{10!}{4! times 6!} = 210`

So the proportion of all possible groups containing the eldest boy is:

`frac{84}{210}=2/5=40%`

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