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April 21, 2014

April 21, 2014

Posted by **Danielle** on Monday, October 7, 2013 at 12:01am.

Please help i got a is 37820 but really need help with part b

(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 -
**Reiny**, Monday, October 7, 2013 at 12:23ama) 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 -
**Danielle**, Monday, October 7, 2013 at 12:26amI understand what you did but which one would i round it to the 20 speeches or 21 speeches?

- Math -
**Reiny**, Monday, October 7, 2013 at 12:36amWell, 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.

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