# Math

posted by on .

A speaker has a collection of 62 jokes. He uses 3 jokes in each speech.

(a) In how many ways can he select 3 jokes for a speech?

How many speeches can he give without using a joke twice? (Round your answer down to the nearest whole number.)

• Math - ,

a) is correct

b) If I interpret it correctly

first speech: he has 62 jokes available
second speech: he has 59 jokes available
third speech: 56 jokes
....
nth speech: ≥ 3

so we have to find the number of terms in the arithmetic sequence
62, 59, 56 , .... , 3 or 2 or 1
a = 62, d=-3 , n = ? when term(n) = 3
a
term(n) = a+(n-1)d
3 = 62 + (n-1)(-3
-59 = -3n + 3
3n = 62
n = 62/3 = 20.666..

check:
term(21) = 62 -3(20) = 2
term(20) = 62 - 3(19) = 5

So he should be able to give 20 speeches, but on the 21st speech he would only have 2 jokes to tell

• Math - ,

I understand what you did but which one would i round it to the 20 speeches or 21 speeches?

• Math - ,

Well, your instructions were to round down to the nearest whole number.
As I said, for his 20th speech he would have 5 jokes left that he hasn't used yet, and he chooses 3 of those.
So on his 21st speech he only has 2 unused jokes, and would have to repeat one of the other jokes.

So without repeating any jokes, my answer would be 20 speeches.