A small acting club has
6
members. Two of the members are to be chosen for a trip to see a Broadway play. How many different
2
-member groups are possible?
what is 6!/(2!4!)
and another way to consider it.
there are six folks for position A, and then 5 to consider for B. So 30 ways to get AB. But BA is the same group, so 30/2 is the number of different member groups.
To find the number of different 2-member groups that can be chosen from a set of 6 members, we can use the concept of combinations.
The formula for combinations is given by:
C(n, r) = n! / (r! * (n-r)!)
where n is the total number of items and r is the number of items to be chosen.
In this case, there are 6 members and we want to choose 2-member groups. So we have:
C(6, 2) = 6! / (2! * (6-2)!)
= 6! / (2! * 4!)
= (6 * 5 * 4!) / (2! * 4!)
= (6 * 5) / (2!)
= 30 / 2
= 15
Therefore, there are 15 different 2-member groups possible from a set of 6 members.
To find the number of different 2-member groups possible from a club of 6 members, you can use the combination formula.
The combination formula is given by:
C(n, r) = n! / ((n - r)! * r!)
where n is the total number of members in the club and r is the number of members to be chosen.
In this case, n = 6 (total number of members in the club) and r = 2 (number of members to be chosen for the trip).
Plugging the values into the formula:
C(6, 2) = 6! / ((6 - 2)! * 2!)
Simplifying further:
C(6, 2) = 6! / (4! * 2!)
Now, let's calculate the factorials:
6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
4! = 4 * 3 * 2 * 1 = 24
2! = 2 * 1 = 2
Substituting the factorials back into the formula:
C(6, 2) = 720 / (24 * 2)
Calculating the division:
C(6, 2) = 720 / 48
Cancelling out the common factors:
C(6, 2) = 15
Therefore, there are 15 different 2-member groups possible from the acting club.