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\(\large \color{black}{\begin{align} & \normalsize \text{The number of ways of dividing 20 different balloons }\hspace{.33em}\\~\\ & \normalsize \text{in two groups containing 11 and 9 balloons is ? }\hspace{.33em}\\~\\ \end{align}}\)
you choose 11 balloons for the first group the other group is automatically determined
i agree with ganeshie8
is answer 20C11
Yes, notice, that also equals 20C9 and agrees with our present problem that you could have chosen the balloons for the second group first
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answer given is 20C11*9C9
20C11 looks correct to me! textbook answer could be wrong, doesn't make sense to m
but 9C9=1
lol okay i read it as 20C11*20C9 for some reason
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