Question Video: Finding the Number of Ways to Choose 𝑛 Out of 𝑚 Things | Nagwa Question Video: Finding the Number of Ways to Choose 𝑛 Out of 𝑚 Things | Nagwa

# Question Video: Finding the Number of Ways to Choose π Out of π Things Mathematics • Third Year of Secondary School

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Miaβs teacher divided the class into groups of 6 and required each member of the group to meet every other member of the same group. How many meetings will each group have?

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### Video Transcript

Miaβs teacher divided the class into groups of six and required each member of the group to meet every other member of the same group. How many meetings will each group have?

In this question, we are told that Miaβs class has been divided into groups of six students. We are told that every member of each group of six must meet with all of the other members. We need to determine how many meetings will take place inside each group of six. There are two different ways we can answer this question. We can start by noting that a meeting in this context means we choose any two students from the same group, that is, the two members of each meeting. In particular, we can note that the order of the two students we choose does not matter. For instance, a meeting between Mia and Emma will be the same as a meeting between Emma and Mia.

We can then recall that π choose π is the number of ways of choosing π objects from π distinct objects, provided π is greater than or equal to π and they are both natural numbers. If we substitute π equals six and π equals two into this expression, then we see that six choose two is the number of ways of choosing two objects from six objects. In this case, it will give us the number of ways of choosing two students from any group of six. Since each choice of two students from the six gives us a meeting, this is the number of meetings that each group will need to have.

We can calculate this value by recalling that π choose π is equal to π factorial over π factorial times π minus π factorial. We can substitute π equals six and π equals two into the formula to obtain six factorial over two factorial times six minus two factorial, which simplifies to give four factorial. We can then evaluate this expression by recalling that a factorial of π is the product of all positive integers less than or equal to π. So, six factorial is six times five times four factorial. We can then cancel the shared factor of four factorial in the numerator and denominator to get six times five over two factorial. We can then calculate that two factorial is equal to two. So this expression evaluates to give 15.

It is worth noting that we can answer this question without using combinations at all. First, we can note that each group has six members. To calculate the total number of meetings this group must make, we can note that one student will meet five other members of the group. For the next student, we have already considered their meeting with the first student. So the student must have four more meetings, one with each of the other members. For the third student, we have already considered their meeting with the first two students. So they must make three more meetings, one with each of the other students.

We can keep applying this process. The fourth student will have two more meetings to attend. And the fifth student will have one more meeting. The sixth student will have already met with all five of the other members at this point. The sum of these values is equal to 15. This also highlights a link between combinations and sums. We can see that π choose two is equal to the sum of the positive integers below π. In any case, we have shown that the total number of meetings required in a group will be 15.

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