# Combinations

#### miaellenaylee

##### New member
Hi,

I need some help with these questions.

Staff member

#### miaellenaylee

##### New member

Have you done any work with the parts (a), (b), etc.? If you did - please share your work - so that we know where to begin to help you!!
No, not really. I'm trying to get back into the topic after quite a while and have forgotten how to attempt these type of questions.
Q11a) C O M B I N A T
O I N
I can recognise that there are 3 letters that repeat. So I would assumed that the answer be 8C6.

11b) Simply because the letters are allowed to repeat I would assume the answer to be 11C6 but the correct answer is far more complicated.

For question 14, I don't even know where to start.

#### pka

##### Elite Member
No, not really. I'm trying to get back into the topic after quite a while and have forgotten how to attempt these type of questions.
Q11a) C O M B I N A T O I N
I can recognise that there are 3 letters that repeat. So I would assumed that the answer be 8C6.

11b) Simply because the letters are allowed to repeat I would assume the answer to be 11C6 but the correct answer is far more complicated.
For question 14, I don't even know where to start.
14a) There are $$\displaystyle \dbinom{10}{6}$$ ways to pick the people to to seated at the large table and we have four left to be seated at the smaller table. Then the six can be arranged in $$\displaystyle (6-1)!=5!$$ and the four can be arranged in $$\displaystyle 3!$$ ways.So what is the total.
14b) There are $$\displaystyle \dbinom{8}{4}$$ ways to pick four people to sit with the couple at the large table.
There are $$\displaystyle \dbinom{8}{6}$$ to pick people to sit at the large table not including the couple.
Now you must still arrange the people at the tables. (note it does not say the couple sits together.)

#### miaellenaylee

##### New member
14a) There are $$\displaystyle \dbinom{10}{6}$$ ways to pick the people to to seated at the large table and we have four left to be seated at the smaller table. Then the six can be arranged in $$\displaystyle (6-1)!=5!$$ and the four can be arranged in $$\displaystyle 3!$$ ways.So what is the total.
14b) There are $$\displaystyle \dbinom{8}{4}$$ ways to pick four people to sit with the couple at the large table.
There are $$\displaystyle \dbinom{8}{6}$$ to pick people to sit at the large table not including the couple.
Now you must still arrange the people at the tables. (note it does not say the couple sits together.)

#### pka

##### Elite Member
You are correct with part a) of #11/ $$\displaystyle \bf{\dbinom{8}{6}}=28$$. SEE HERE

There is not standard, clever way to do part b), we will just take it apart piece by piece. That is tedious in the extreme.

There is an advanced way if one understands generating functions: $$\displaystyle {(1 + x)^5}{(1 + x + {x^2})^3}$$, SEE THE EXPANSION
In that expansion the term $$\displaystyle 179x^6$$ tells us that $$\displaystyle 179$$ is the answer.

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#### miaellenaylee

##### New member
you are correct with part a) of #11/ $$\displaystyle \bf{\dbinom{8}{6}}=28$$. see here

there is not standard, clever way to do part b), we will just take it apart piece by piece. That is tedious in the extreme.

There is an advanced way if one understands generating functions: $$\displaystyle {(1 + x)^5}{(1 + x + {x^2})^3}$$, see the expansion
in that expansion the term $$\displaystyle 179x^6$$ tells us that $$\displaystyle 179$$ is the answer.
thank you

#### HT1992

##### New member
Probability of a range of numbers occurring in a combination

i am not a mathematician or a math student but i have come upon this problem in my work. i want to for example if i am to choose 5 numbers from 1 to 50 what is the probability of at least one of them being any of the specific 10 numbers of the total 50 (for example 41 to 50)?

sorry if its really basic but i really like to know this thank you in advance!