A speaker has a collection of 62 jokes. He uses 3 jokes in each speech.
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.)
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
I understand what you did but which one would i round it to the 20 speeches or 21 speeches?
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.
what did the evil chicken lay
To find the number of ways the speaker can select 3 jokes for a speech, we can use the combination formula, also known as "n choose r". The formula is given by:
C(n, r) = n! / (r! * (n - r)!)
where "n" represents the total number of items (in this case, jokes), and "r" represents the number of items to be chosen (in this case, 3 jokes).
For part (a), the speaker wants to select 3 jokes for a speech, so n = 62 and r = 3.
Using the formula, we can calculate the number of ways as follows:
C(62, 3) = 62! / (3! * (62 - 3)!)
Now, let's simplify the equation and calculate the answer:
C(62, 3) = 62! / (3! * 59!)
Here's how you can calculate the answer:
Step 1: Calculate 62! (the factorial of 62) by multiplying all the numbers from 1 to 62.
Step 2: Calculate 3! (the factorial of 3) by multiplying all the numbers from 1 to 3.
Step 3: Calculate 59! (the factorial of 59) by multiplying all the numbers from 1 to 59.
Now, plug these values into the formula:
C(62, 3) = 62! / (3! * 59!) = (1 * 2 * 3 * ... * 62) / ((1 * 2 * 3) * (1 * 2 * ... * 59))
By simplifying the expression, we can cancel out common factors:
C(62, 3) = (60 * 61 * 62) / 6 = 238,760
Therefore, there are 238,760 ways the speaker can select 3 jokes for a speech.
Moving on to part (b), we need to find out the number of speeches the speaker can give without using a joke twice.
Since a speech requires 3 jokes and there are 62 jokes in total, the speaker can give a maximum of floor(62/3) = 20 speeches, where "floor" is the round-down function.
Hence, the speaker can give a maximum of 20 speeches without using a joke twice.