1
$\begingroup$

Fifty-five identical red balls and three identical green balls are to be distributed among seven children. Each child must get at least five balls. In how many ways can this be done?

What I have so far:

Well, I am guessing that we can figure out the case for the number of ways that just the red balls can be distributed among the children. Since 35 of the balls must already be given to the children, there are 20 balls left over so the number of ways that just the red balls can be distributed is ${20}\choose{7}$. And then you have to figure out the number of ways that the red and green balls combined can be distributed among the children. The three green balls can be distributed among seven children as $3\choose7$. And then you find the number of ways that the left over red balls can be distributed amont these children when some of them have green balls. Is this correct?

$\endgroup$
2
  • $\begingroup$ Note that $\binom{3}{7} = 0$. Is that what you actually meant? $\endgroup$ Apr 16, 2015 at 2:30
  • $\begingroup$ Yes. You are talking to the village idiot here. $\endgroup$
    – EmaLee
    Apr 16, 2015 at 2:42

3 Answers 3

6
$\begingroup$

HINT: If we were just distributing the $55$ red balls, the answer would be $\binom{20+7-1}{7-1}=\binom{26}6$, not $\binom{20}7$: this is a straightforward stars and bars problem, and the reasoning justifying this calculation is described fairly clearly at the link. (Your realization that we're effectively distributing only $20$ balls is correct, however.)

To solve the actual problem, I would first pretend that all $58$ balls are indistinguishable; the same reasoning would then give you $\binom{23+7-1}{7-1}=\binom{29}6$ possible distributions. Now we have to determine in how many distinguishable ways we can turn $3$ of the $58$ balls red. Each child has more than $3$ balls, so choosing $3$ balls to turn red is basically the same as choosing a way to distribute $3$ indistinguishable balls among $7$ people. Using the stars and bars calculation yields not $\binom37$, but rather ... ?

Now what must you do with that last number and the $\binom{29}6$ in order to get the total number of possibilities?

$\endgroup$
6
  • $\begingroup$ Okay, so to distribute 3 balls to 7 people it is ${3+7-1}\choose{7-1}$? And then you just multiply those two numbers together to get the total number? $\endgroup$
    – EmaLee
    Apr 16, 2015 at 2:55
  • $\begingroup$ I think this is the way to go, no case work require, I would like to point out that no case work is needed is because each person has at least five balls, which is larger than three. So we can use stars and bars for a second time without having to put artificial upper bounds on the number of green balls per person, which would make the problem a lot more tedious. $\endgroup$
    – Asinomás
    Apr 16, 2015 at 2:56
  • $\begingroup$ Also when do you know that the combination is ${n+k-1}\choose{k-1}$ as opposed to just ${n}\choose{k}$ ? $\endgroup$
    – EmaLee
    Apr 16, 2015 at 2:58
  • $\begingroup$ @EmaLee: To answer your first comment: yes to both. For the second comment, did you read the explanation at the stars and bars link? $\endgroup$ Apr 16, 2015 at 3:01
  • $\begingroup$ No, but I will do that. Thank you. $\endgroup$
    – EmaLee
    Apr 16, 2015 at 3:01
4
$\begingroup$

Since $3$ is a small number, we can divide into cases.

Case (i) One child gets all the greens.

Case (ii) One child gets $2$ greens, and another gets $1$.

Case (iii) Three children get $1$ green each,

We analyze Case (i) in fair detail, and leave the other two cases to you.

The child who gets the greens can be chosen in $\binom{7}{1}$ ways. Give her $2$ reds, and give the others $5$ reds each. So $32$ reds are gone. That leaves $23$ reds, which can be distributed among the $7$ children in $\binom{23+7-1}{7-1}$ ways (Stars and Bars, please see Wikipedia). Thus there are $\binom{7}{1}\binom{29}{6}$ ways to do the distribution so that one child gets all the greens.

$\endgroup$
0
$\begingroup$

firstly we should distribute $3$green balls to $7$ children using $\binom{n+r-1} {r-1}$ where n is number of objects and r is number of groups this formula gives us $\binom{9}{3}$ then we should add $32$ red balls to make all groups have 5 balls then we will distribute last 23 balls using the same formula so we should multiply $\binom{9}{3}$ with $\binom{29}{6}$, it is a very big number so I will leave it as it is.$$\binom{9}{3}\times\binom{29}{6}$$

$\endgroup$

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.