There are 10 marbles in a bag. $6$ are red and $4$ are blue. You must chose at least 1 red marble. In how many ways can you chose three total marbles.

Question

I thought the answer is $^9C_2$ since the first (red) marble didn't count. You have to pick a red marble which reduces the total count from 10 to 9. The answer is 116 possible ways.

Answer 1

There are total of $10$ balls in which $6$ are red and $4$ are blue.

First you can choose $1$ Red and $2$ blue in $\dbinom{6}{1}\dbinom{4}{2}=36$ ways

Second you can choose $2$ Red and $1$ blue in $\dbinom{6}{2}\dbinom{4}{1}=60$ ways

Third you can choose all $3$ red in $\dbinom{6}{3}=20$ ways

In total you can choose in $\dbinom{6}{1}\dbinom{4}{2}+\dbinom{6}{2}\dbinom{4}{1}+\dbinom{6}{3}=36+60+20=116$ ways

Answer 2

Hint: In how many ways can you choose three marbles, none of which are red?

Answer 3

The overall numbers of ways to choose 3 marbles are

$$N_1=\binom{10}{3}=120$$

The overall numbers of ways to choose 3 blue marbles are

$$N_2=\binom{4}{3}=\binom{4}{1}=4$$

therefore $N=N_1-N_2=116$.

Answer 4

you want to choose three marbles and must choose at least a red marble.

Here are the ways in you can do that

1Red 2Blue

2Red 1Blue

3Red

$$total=\binom{6}{1}\binom{4}{2}+\binom{6}{2}\binom{4}{1}+\binom{6}{3}=116 ways$$