# Permutations / probability question

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• Mar 10th 2009, 11:24 AM
champrock
Permutations / probability question
• Mar 10th 2009, 02:29 PM
Plato
There are ${10 \choose 3}$ ways to pick the numbers to match. That is six cards.

So we need four more cards. They can be of either color, but not the same number.
Therefore, there are ${7 \choose 4}$ ways to pick the numbers to not match.
But those four cards can be of either color, that is $2^4$ ways.

What is the correct answer?
• Mar 11th 2009, 04:24 PM
Soroban
Hello, champrock!

I think I've got it . . .

Quote:

A box contains 10 red cards number from 1 to 10
and 10 black cards numbered from 1 to 10.

If 10 of the 20 cards are chosen, in how many ways can there be exactly three matches?
(A match means a red card and a black card with the same number.)

. . $(A)\;{10\choose3}{7\choose4}2^4 \qquad (B)\;{10\choose3}{7\choose4} \qquad (C)\;{10\choose3}2^7 \qquad (D)\;{10\choose3}{14\choose4}$

We choose three of the ten red cards . . . There are: ${\color{blue}{10\choose3}}$ ways.

We want the three matching black cards . . . There is: $1$ way.

Now we want four more cards that do not match each other.
Choose four of the seven remaining red cards . . . There are: . ${\color{blue}{7\choose4}}$ ways.
. . But each of those red cards can be exchanged for its black mate.
. . There are: . ${\color{blue}2^4}$ possibilities.

Therefore, there are: . ${10\choose3}{7\choose4}2^4$ ways . . . answer (A).

Awww, Plato beat me to it . . .
.
• Mar 12th 2009, 04:20 AM
champrock
hmm

till selecting 10c3 red balls is fine and there is 1 combination for corresponding combination for balck balls.

But after that, there are we have got 14balls out of which we have to choose 7. So, what is the problem if we just use normal combinations to this ?? (select 7 out of 14) ?

Dont understand why we have to do 2^4