What is the probability of getting 2 same colour sweets and 1 different colour sweet? A little box contains $40$ smarties: $16$ yellow, $14$ red and $10$ orange.
You draw $3$ smarties at random (without replacement) from the box.
What is the probability (in percentage) that you get $2$ smarties of one color and another smarties of a different color?
Round your answer to the nearest integer.
Answer given is $67$. I don't get it. Is it not:
$$\left(\frac{16}{40} \times \frac{15}{39} \times\frac{24}{38}\right) + \left(\frac{14}{40} \times\frac{13}{39} \times\frac{26}{38}\right) +\left(\frac{10}{40} \times\frac{9}{39} \times\frac{30}{38}\right)= 22?$$

We imagine taking out the candies one at a time.
Your $\frac{16}{40}\cdot\frac{15}{39}\cdot \frac{24}{38}$ calculates the provability of getting Yellow, Yellow, Other in that order. However, two Yellow and one Other can happen in two additional orders, Yellow, Other, Yellow or Other, Yellow, Yellow. Each of these turns out to have the same probability as Yellow, Yellow, Other.
So the probability of $2$ Yellow and $1$ Other is $3\cdot \frac{16}{40}\cdot\frac{15}{39}\cdot \frac{24}{38}$.
Similar adjustments need to be made in your other two terms.
Another way: There are $\binom{40}{3}$ equally likely ways to choose $3$ candies. We now count the favourables, where we have $2$ candies of one colour and $1$ of another. For example the number of ways to have $2$ Yellow and $1$ other is $\binom{16}{2}\binom{24}{1}$, and we have similar expressions for the other favourables. Add up, and divide by $\binom{40}{3}$.
