A grandmother has 7 grandchildren, and 14 apples to give. How many ways can she give apples to her grandchildren so that each grandchild gets aT LEAST one? (but she has to get rid of hers).

This seems like a stars and bars? My guess:

How many ways she gave = $\displaystyle \binom{13}{6}$

  • $\begingroup$ The answer is correct, the problem has one of the two standard "Stars and Bars" shapes. $\endgroup$ Feb 4, 2015 at 12:31
  • $\begingroup$ @AndreNicolas, issue. I used: $\binom{14 - 1}{7 - 1}$ method. I did NOT use the method described by axiom. $\endgroup$
    – Lebes
    Feb 4, 2015 at 12:43
  • $\begingroup$ Yes, you found the number of solutions of the equation in positive integers directly, instead of travelling through the non-negative solutions of $x_1+\cdots+x_7=14-7$. So your way is more efficient, one visualizes lining up the apples and choosing $6$ "gaps" to put a separator into. $\endgroup$ Feb 4, 2015 at 12:47

2 Answers 2


Let us give one Apple to each of the child (as per the constraint), which leaves us with 7 Apples and 7 kids.

The number of ways in which you can distribute apples thus becomes the number of solutions to

$Kid_1 + Kid_2 + ... + Kid_7 = 7$

where $Kid_i$ is the number of apples given to the $i^{th}$ kid.

This is $\binom{7 - 1 + 7}{7} = \binom{13}{7} = \binom{13}{6}$, so your answer is correct.

  • $\begingroup$ +1 this is how I would do it. It would be nice if you included an explanation of why the number of solution to that equation is ${7 - 1 + 7 \choose 7}$ :-) $\endgroup$
    – Ant
    Feb 4, 2015 at 12:35

1) All apples are identical

2) All kids are distinguishable

This is a generalization for $n$ apples

Write n as $1+1+ \ldots +1$. If we are to split between 2 kids, we obvjously need only 1 bar between 1's and we have n+1 positions for it, so for 7 we need 6 bars with a total of $n+6$ positions (just write $1+1+\ldots 1$+bar1+bar2+...+bar6 - a total of n+6 positions).

Since you have no more constraints for splitting the number of apples among kids, you can put these 6 bars anywhere in n+6 postitions, giving you a total of $\binom{n+6}{6}$ ways of doing this.

  • $\begingroup$ To the kids, of course, the apples are easily distinguishable - someone else always has a better apple than you. $\endgroup$
    – Joffan
    Feb 4, 2015 at 21:40
  • 1
    $\begingroup$ I guess it's called a model for a reason $\endgroup$
    – Alex
    Feb 4, 2015 at 22:08

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