Your math club has 20 members. In how many ways can it select a president and a vice-president if no member can hold more than one office?
+2094 The formula is for combinations: \(\frac{n!}{k!(n-k)!}\)
Plugging in, we get
20!/(2! times 18!) which will get you 190.
So, the answer is 190 ways.
+128475 We can choose any 1 of the 20 members for president
Once any of these is selected, we can choose any 1of the remaining 19 members for vice-president
So .....by the fundamental counting principle = 20 * 19 = 380 ways
+2094 Chris, I don't get how my answer is incorrect. Will you mind explaining why it is wrong?
+128475 OK, Cal....you ask a good question....!!!!
Let's look at something more simple
Suppose we only had 4 members A, B C D
Let A/B A/C A/D B/C B/D C/D represent president / vice-president
Note that this is C(4,2) = 4!
_______ = 6
(4 - 2)! 2!
But....what we have as actually a permutation because A/B isn't the same as B /A
So we actually have P(4,2) = 12
So your answer is a combination C(20,2) = 190
But we should have a permutation = P(20, 2) = 380...... this is the same result as 20 * 19
