# combinations and permutations - choosing when there's a limit

1. A kid can choose 7 out of 12 donuts to eat. How many ways can he do this if he must choose at exactly 3 of the first 5?

2. Similarly, how many combinations are there if he must choose at least 3 of the first 5.

The way I approached this is 5C3*9C4 if he must choose exactly 3 of the first 5, but for the second part where he must choose at least 3 of the first 5, I have 5C3*9C4+5C4*8C3+5C5*7C2. Is my reasoning correct?

For the first question, note that it is $\binom{5}{3}\binom{7}{4}$. (I am assuming the doughnuts are lined up in a row.)

For there is a total of $12$ doughnuts. If we are to eat $7$, exactly $3$ of which are among the first $5$, we must choose $3$ from the first $5$, and the remaining $4$ from the last $7$.

For at least three, it is the same idea. Add to our expression for exactly three the number $\binom{5}{4}\binom{7}{3}$ for exactly four, and $\binom{5}{5}\binom{7}{2}$ for exactly five.

• Thank you for the feedback. But why is it 7C4 instead of 9? Aren't there 9 remaining options after he's already chosen 3? This is of course if choosing 3 from the first 5 doesn't disqualify the remaining 2 from being chosen (which i didnt specify I apologize) – sofreakinlost Jun 3 '14 at 4:38
• If the kid must eat exactly $3$ from the first $5$, she cannot choose more than $3$ from these, even if the flavours look tempting. – André Nicolas Jun 3 '14 at 4:40
• oh i see now thank you! – sofreakinlost Jun 3 '14 at 4:42
• You are welcome. Parenthetically, it is tempting to use $\binom{5}{3}\binom{9}{4}$ for the second question (at least $3$). But you did not yield to temptation. It turns out that $\binom{5}{3}\binom{9}{4}$ counts more than once some choices of at least $3$ from the first group. – André Nicolas Jun 3 '14 at 4:47

For the first it should be: $\displaystyle\large{^5C_3}\cdot{^{12-\color{red}{5}}C_{7-3}}$.

That is, subtract the number of "good" choices from the total number of choices. Which is to say, after choosing 3 from the first 5, the goat must choose the remaining 4 from the last 7.

For the second, you have the correct logic (add the ways to get exactly 3, 4, and 5), but made the same mistake.