A bag 2

A bag contains seven green and eight red jellybeans. How many ways can five jellybeans be withdrawn from the bag so that the number of green ones withdrawn will be less than 4?

Correct answer:

n =  26

Step-by-step explanation:

$$\begin{gathered} 0+5 \\ 1+4 \\ 2+3 \\ 3+2 \\ {C}_0(5)=\left(\genfrac{}{}{0ex}{}{5}{0}\right)=\frac{5!}{0!(5-0)!}=\frac{1}{1}=1 \\ {n}_1=\left(\genfrac{}{}{0ex}{}{5}{0}\right)=1 \\ {C}_1(5)=\left(\genfrac{}{}{0ex}{}{5}{1}\right)=\frac{5!}{1!(5-1)!}=\frac{5}{1}=5 \\ {n}_2=\left(\genfrac{}{}{0ex}{}{5}{1}\right)=5 \\ {C}_2(5)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\frac{5!}{2!(5-2)!}=\frac{5\cdot 4}{2\cdot 1}=10 \\ {n}_3=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10 \\ {C}_3(5)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!(5-3)!}=\frac{5\cdot 4}{2\cdot 1}=10 \\ {n}_4=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10 \\ n=n_1+n_2+n_3+n_4=1+5+10+10=26 \end{gathered}$$

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