1
$\begingroup$
  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?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – sofreakinlost Jun 3 '14 at 4:38
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Jun 3 '14 at 4:40
  • $\begingroup$ oh i see now thank you! $\endgroup$ – sofreakinlost Jun 3 '14 at 4:42
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Jun 3 '14 at 4:47
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.