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Find the total number of ways to select at least one red ball from a bag containing $4$ red balls and $3$ green balls, balls of the same colour being identical, if any number of balls may be selected.

For each selection of at least one red ball, green balls can be selected in $4$ ways, i.e. $\{0,1,2,3\}$ number of green balls selected. The answer should be $4 \times 4 = 16$.

But the answer given is $8$.
What am I missing?

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  • $\begingroup$ Not following. there are $4$ ways to select a red ball as there are $4$ red balls. Or there's only one way, as the red balls are all the same. Are you selecting multiple balls? Is there replacement? $\endgroup$
    – lulu
    Mar 25, 2016 at 13:17
  • $\begingroup$ There are four ways beacuse any number of balls can be selected. $\endgroup$
    – mathemather
    Mar 25, 2016 at 13:21
  • $\begingroup$ Ok, then I agree with $16$. Easy to list them all...can't see any argument for ignoring some of them. $\endgroup$
    – lulu
    Mar 25, 2016 at 13:26

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Your answer is right. If the selections are distinguished according to the number of red balls and the number of green balls they contain, and all selections except those with $0$ red balls are to be counted, there are $4\times4=16$ different selections.

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