0
$\begingroup$

You have $20$ cards and $12$ envelopes(labelled $1.....12$). In how many ways can you put $20$ cards in envelopes if

a) the cards are distinct

b) the cards are identical

c) the cards are identical and no envelope can be left empty.

Please provide me with some briefing on how to solve this problem. Thank you

$\endgroup$
  • $\begingroup$ Can you put more than one card into each envelope? $\endgroup$ – stuart stevenson Mar 5 '19 at 12:07
  • $\begingroup$ Yes you can. You have to fit all 20 cards into 12 envelopes $\endgroup$ – Sara Rafiq Mar 5 '19 at 12:09
  • 1
    $\begingroup$ If you're stuck, it can often be a good idea to first try simpler problems. How about $20$ cards and $2$ envelopes? Or $2$ cards and $12$ envelopes (or $14$ cards and $12$ envelopes for c). Then generalize from there. Or if you can't generallize, take it one step further: $20$ cards and $3$ envelopes. Do you see any patterns that might be generalizable? If you're still stuck, that's a great time to tell us what answers you got and how you got them, and we will me much better suited to help you. $\endgroup$ – Arthur Mar 5 '19 at 12:23
  • $\begingroup$ What have you tried so far? $\endgroup$ – Vinyl_cape_jawa Mar 5 '19 at 12:23
1
$\begingroup$

HINTS:

For a) think how many different places/envelopes can card nr $1$ end up in? What is the situation with card nr $2$? You may found useful the rule of product.

For b) you could read up on stars and bars.

For c) start by fulfilling the condition and put one card in each envelope. How many cards do you have left? Which of a) or b) is to be used with the leftover cards?

Hope this helped

$\endgroup$
2
$\begingroup$

a) for each of the $20$ cards there are $12$ possibilities, so...

b) To be found is the number of sums $c_1+\cdots+c_{12}=20$ where the $c_i$ are nonnegative integers. For this you can apply stars and bars.

c) To be found is the number of sums $c_1+\cdots+c_{12}=20$ where the $c_i$ are positive integers.

Define $c'_i:=c_i-1$ and find the number of sums $c'_1+\cdots+c'_{12}=20-12=8$ where the $c'_i$ are nonnegative integers, again with stars and bars.

$\endgroup$
-1
$\begingroup$

a) $$ \frac{31!}{11!} $$

A example:

$$ 1, 3, 5 | 2 | 4 | | | 6, 7, 8 | | | | | | 9, 10, 11, 12, ..., 20 $$

31 Symbols (20 numbers + 11 separators). The 11 separators repeating. The separators, separating the envelopes.

b) $$ \frac{31!}{11! \cdot 20!} $$

A example:

$$ A, A, A | A | A | | | A, A, A | | | | | | A, A, A, A, ..., A $$

31 Symbols (20 'A' + 11 separators). The 11 separators repeating. The separators, separating the envelopes.

c) $$ \frac{19!}{11! \cdot 8!} $$

A example:

$$ A, A, A | A | A | A | A | A, A, A | A | A | A | A | A | A, A, A, A, A $$

31 Symbols (20 'A' + 11 separators). The 11 separators repeating. The separators, separating the envelopes.

In this example we take an A from each envelope and it remains:

$$ A, A | | | | | A, A | | | | | | A, A, A, A $$

19 Symbols (8 'A' + 11 separators). The 11 separators repeating. The separators, separating the envelopes.

$\endgroup$
  • $\begingroup$ I think you have a) wrong. But it's well laid out. I'll upvote when you correct. $\endgroup$ – stuart stevenson Mar 5 '19 at 15:51

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.