+0

Help is appreciated

+2
306
4
+165

Our club has 25 members, and wishes to pick a president, secretary, and treasurer. In how many ways can we choose the officers, if individual members are allowed to hold 2 but not all 3 offices?

(if you give an answer always explain it, answers mean nothing compared to an understanding of the problem)

May 1, 2021

#1
+201
+1

Okay I will try to explain this as best as I can.

If the members were not allowed to hold two positions, the answer would be 25*24*23.

This is because there are 25 people who can run for the first spot, and 24 people who can run for the second, because

the first spot would be taken, and there will only be 24 people availible for the second. Same goes with the 23. (I hope I am not making this confusing.)

But because we can't do that, we have to try something else. If a member can hold two spots, the number of spots will be

25*25*24. This accomendates for two spots and allows a member to have two positions. 25*25*24 is equal to 15,000

So there are 15,000 combinations.

May 2, 2021
edited by mathisopandcool  May 2, 2021
#2
+201
+2

and tell me how I can make my explanations better

:)

mathisopandcool  May 2, 2021
#3
+2266
-1

Nice explanation :))

But shouldn't it be 25*24*3 instead of 25*25*24?

25*24 ways to choose 2 people, and 3 ways to then assign those 2 people.

=^._.^=

catmg  May 2, 2021
#4
+201
-1

O ya ur right im bad at math lmao srry

mathisopandcool  May 2, 2021