# How many combinations contain specific numbers?

• Jan 22nd 2011, 06:46 AM
How many combinations contain specific numbers?
Given are all possible combinations of 5 numbers from 1-35, all contained in combinations of 6 numbers (from 1-35). How many of the six number combinations contain 5 specific numbers?
• Jan 22nd 2011, 07:07 AM
Plato
Quote:

Given are all possible combinations of 5 numbers from 1-35, all contained in combinations of 6 numbers (from 1-35). How many of the six number combinations contain 5 specific numbers?

Is it possible to give a clearer statement of this question?
As it reads now, you are asking “how many combinations of six contain a particular combination of given numbers?” The answer to that is 30.
But I cannot imagine that is what the question means. Is it?
• Jan 22nd 2011, 07:17 AM
Well, I'm a bit confused! That's what I thought at first too.

Since we are dealing with six number combinations, each one of which contains 6 possible combinations of five numbers, the total number of combinations of six containing all the combinations of five would be 324632 (35c5) / 6 = 54105 combinations of six; However, 35c6 would be 1623160.

So, if 30 were the answer to the first option, what would happen if we had all the combinations of 6 instead? Would it just be the same?
• Jan 22nd 2011, 07:23 AM
Plato
I just do not know what the OP means.
Is that an exact wording or is it a translation?
• Jan 22nd 2011, 07:36 AM
It's a translation. :)

Let me paraphrase.

We have all the possible combinations of five numbers from 1 to 35 (35c5) contained in combinations of six numbers.
for example: 1 2 3 4 5 6 (contains {1,2,3,4,5} {1,2,3,4,6} {1,2,3,5,6} {1,2,4,5,6} {1,3,4,5,6} {2,3,4,5,6})

What I want to know is how many of these -six- number combinations contain -five- specific numbers. Logically its 30. I'm confused however with the fact that every >1< combination of 6 contains 6 combinations of five!! :)
• Jan 22nd 2011, 07:46 AM
Plato
There are six combinations of six objects taken five at a time.
$\displaystyle \dbinom{6}{5}=\dfrac{6!}{5!\cdot 1!}=6$
• Jan 22nd 2011, 08:40 AM
Soroban

The wording is confusing.
All that emphasis on subsets of 5 numbers . . .

Quote:

Given are all possible 6-number subsets of numbers from 1-35,
how many of these subsets contain 5 specific numbers?

What do you mean by "logically it's 30"? .Is that the correct answer?

Here's my reasoning . . .

The is one way to get the five specific numbers.
The sixth number can be any of the remaining 30 numbers.