Combinations in combinations?

**aimpro2000** · September 19th 2010, 02:57 PM

Given are all the possible 10 number combinations (NOT permutations) of the numbers 11 - 26 (8008 combinations), and 4 random numbers from the 11 - 26 sequence.
How do I find out how many of the 8008 combinations contain
a. all 4 random numbers?
b. 3 of the 4, 2 of the 4, 1 of the 4 and 0 of the 4 random numbers respectively??

**Plato** · September 19th 2010, 03:23 PM

Originally Posted by aimpro2000

Given are all the possible 10 number combinations (NOT permutations) of the numbers 11 - 26 (8008 combinations), and 4 random numbers from the 11 - 26 sequence.
How do I find out how many of the 8008 combinations contain
a. all 4 random numbers?
b. 3 of the 4, 2 of the 4, 1 of the 4 and 0 of the 4 random numbers respectively??

What do you mean by random number?

**aimpro2000** · September 19th 2010, 03:27 PM

any 4 numbers from 11 -26!

**Plato** · September 19th 2010, 03:33 PM

Sorry, in that case your question makes no sense whatsoever.
Try to restate the question is a understandable form.

**aimpro2000** · September 19th 2010, 03:43 PM

I'm sorry, maybe that "random" is confusing. Let's say I need to find out how many combinations of 10 numbers from 11 - 26 contain the numbers 11, 12, 13, 14!!

**Plato** · September 19th 2010, 03:57 PM

Originally Posted by aimpro2000

I'm sorry, maybe that "random" is confusing. Let's say I need to find out how many combinations of 10 numbers from 11 - 26 contain the numbers 11, 12, 13, 14!!

The answer to that particular question is: $\dbinom{12}{6}$ .

**aimpro2000** · September 19th 2010, 04:01 PM

thanks

**Archie Meade** · September 19th 2010, 05:04 PM

Originally Posted by aimpro2000

Given are all the possible 10 number combinations (NOT permutations) of the numbers 11 - 26 (8008 combinations), and 4 random numbers from the 11 - 26 sequence.
How do I find out how many of the 8008 combinations contain
a. all 4 random numbers?
b. 3 of the 4, 2 of the 4, 1 of the 4 and 0 of the 4 random numbers respectively??

(a)

There are 16 numbers from 11 to 26 inclusive.
If you set aside your four random numbers, you want to choose 6 of the remaining 12 to go with them.

There are $\displaystyle\binom{16-4}{10-4}$ ways to do that.

(b)

There are $\displaystyle\binom{4}{3}$ ways to choose 3 of the 4 random numbers.

For each of those choices there are $\displaystyle\binom{16-4}{10-3}$ ways to choose 7 numbers to go with them,
such that 3 of the 4 random numbers are chosen.

There are $\displaystyle\binom{4}{2}$ ways to choose 2 of the 4 random numbers.

There are $\displaystyle\binom{16-4}{10-2}$ ways to match them with 8 others that do not belong to the random numbers.

Same strategy for the remaining selections.

**mr fantastic** · September 19th 2010, 10:21 PM

Originally Posted by aimpro2000

any 4 numbers from 11 -26!

*ahem* 26! is quite a large number ....Perhaps you just mean 26.

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