4
$\begingroup$

This is a problem I thought of while learning combinatorics, so it may be missing something.

You have 11 socks (5 red socks and 6 blue socks) and 4 drawers. What is the number of ways that you can distribute 11 socks into 4 drawers?

Now when I do this I thought of using the following formula $$ x_1 + x_1' + x_2 + x_2' + x_3 + x_3' + x_4 + x_4' = 11 $$ with the primed ones being the number of blue socks in each drawer (there are four, hence the numbering) and the others being the number of red socks in each drawer. Then I impose the following conditions: $ x_1, x_2, x_3, x_4 \leq 5 $ and $ x_1' ,x_2', x_3', x_4' \leq 6 $.

But then I remember that $ x_1 + x_2 + x_3 + x_4 = 5 $ and $ x_1' + x_2' + x_3' + x_4' = 6 $. There is a formula that counts the number of ways to distribute k objects into n boxes, which can be used in this situation: $$ \binom{n + k - 1}{k} $$

By the product rule I can just arrive at the solution: $$ \binom{4 + 5 - 1}{5} \cdot \binom{4 + 6 - 1}{6} = 4704 $$

So is the problem well-defined? Is the solution above correct? Is there a simpler solution?

$\endgroup$
0
$\begingroup$

Split the problem into two parts, then multiply the results. Distribute the 5 red socks into 4 drawers.

$$ n_R = C(n+k-1,k-1) = C(5+4-1,4-1) = C(8,3) $$

Now distribute the 6 blue socks into 4 drawers.

$$ n_B = C(n+k-1,k-1) = C(6+4-1, 4-1) = C(9,3) $$

The solution should be the product of the two.

$$ n_T = n_R n_B = C(8,3)\cdot C(9,3) = 4704 $$

$\endgroup$
  • $\begingroup$ Shouldn't it be $ C(n + k - 1, k) $? According to this $\endgroup$ – DudeLearningStuffs Nov 29 '16 at 1:47
  • $\begingroup$ You're distributing $n$ indistinguishable objects into $k$ distinguishable boxes, the formula for which is given by $C(n+k-1,k-1)$. $\endgroup$ – Anthony P Nov 29 '16 at 1:52
  • $\begingroup$ I see, but I think you substituted 4 for $ k - 1 $ in the second part of the combinations, while it should be (4 - 1). $\endgroup$ – DudeLearningStuffs Nov 29 '16 at 1:55
  • $\begingroup$ Yes, you're right. Sorry, 4 drawers not 5. $\endgroup$ – Anthony P Nov 29 '16 at 1:56

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.