Count the pairs

Probability Level 3

Eight people are at a party and pair off to form four teams of two. In how many ways can this be done?

8 × 7 × 6 × 5 8\times7\times6\times5 4 × 3 × 2 4\times3\times2 7 × 5 × 3 7\times5\times3 None of these options

Shivdutt Sharma by shivdutt sharma , 29, India

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4 solutions

Shivdutt Sharma
Jan 1, 2017

For first person there are 7 options, for second person has 5 option and so on. 7 x 5 x3 x 1

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Eric Kim
Jan 1, 2017

There are 8×7/2 ways to choose the first group, 6×5/2 for the second, then 4×3/2 and 2×1/2. Multiplying them all together, we get 7×6×5×4×3.

Multiplying them together implies that the order of the teams matters, when it doesn't. To prevent overcounting, we divide by 4!

7×6×5×4×3/(4×3×2) = 7 × 5 × 3

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Why do u divide by 2 for each group

Ashish Sacheti - 4 years, 5 months ago

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Example using the first group: Say you choose person A (8 options) and then person B (7 options). This is why there is a 8*7.

However, you could also choose person B (8 options) and then person A (7 options).

This would give the same result. Since every pairing can be flipped, you must divide by 2.

Alex Li - 4 years, 5 months ago

And why did u do Debby 4!?

Ashish Sacheti - 4 years, 5 months ago

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Lets look at a set of size 2: {A,B}. How many different groups of size 2 can you make?

You have 2 possibilities for the 1st term, and 1 possibilities for the 2nd term. This makes it 2×1 permutations, or AB and BA in this case.

However, in this problem, the order of the terms don't matter. Thus, AB and BA are the same group (This is like a partner project in school. A group of you and Joe is the same as a group of Joe and you).

So you divide the number of ways you can order AB, for that is that number of times you overcounted the number of terms. Finding the number of orderings of distinct terms is easy. It's just 2!, or the factorial of the number of terms. So 2/2! = 1, and the only group possible is (A,B).

Same with this problem. After I multiply the number of ways I can find the 1st, 2nd, 3rd, and 4th terms, I realize that this counts different orderings of the same 4 groups. Thus, I divide by 4!, the number of ways to order 4 groups.

I tried to be as clear as possible, but I'm pretty new to teaching concepts so make sure to tell me if any of it seems off.

Eric Kim - 4 years, 5 months ago

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thank you!

Ashish Sacheti - 4 years, 5 months ago

Sorry *divide by

Ashish Sacheti - 4 years, 5 months ago

We have to make 4 teams.So,the number of options, = 8 c 2.6 c 2.4 c 2.2 c 2 3 ! \frac{8c2 . 6c2 . 4c2 . 2c2}{3!} = 105 = 7 x 5 x 3

General solution of these problems such as divide n things between g groups where every group get m things.

solutions = n ! ( m ! ) g . g ! \frac{n!}{(m!)^g.g!}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Let's say that 4 people are going to select their teammates from the others.

The first person has 7 options for partners.

The next person has 5 other people to choose from because 2 people have already paired off.

The next person has 3 other people to choose from because 4 people have already paired off.

Next person has just 1 person to choose from because 6 people have already paired off.

Number of ways = (7)(5)(3)(1) = 105

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