Question

ey need elp rite now pleaz...
Of 8 households chores, in how many ways can you do three-fourths of them?
Latanya volunteers to bake and deliver pastries to elderly people in her neighborhood. In how many different ways can Latanya deliver to 2 out of 6 elderly people in her neighborhood?

Answer 1

you need to do 6 of the 8

then its 8 x 7 x 6 x 5 x 4 x 3 =

Answer 2

wait..@campbell_st, do you mean the first one?

Answer 3

yep the 1st one

Answer 4

okay, thanks, then its continous, rite?

Answer 5

8!
rite?

Answer 6

This is very simple combinatorics. In the first problem, you want to do 3/4 of the 8 household chores, or 6 total. If the order you do them in matters, campbell's answer is correct. This is because under permutation, doing chore A first and B second would be a different case than doing chore B first and A second. Combination simply looks at the number of different sets regardless of order.

To choose a items from b items total, the notation is a C b, and is given by the formula b!/(a!*[b-a]!) (where n! = n*n-1*n-2*n-3*...*3*2*1)

For a = 6, b = 8, there are 8!/(6!*2!) = 8*7/2 = 28 ways to pick a set of 6 chores to do.


Similarly, in the second problem if you consider visiting Mr. Red then Mr. Blue to be a different case than visiting Mr. Blue then Mr. Red, there are 6x5 = 30 possible ways to visit 2 people. If order doesn't matter, you use combination again: 6!/2!*4! = 6*5/2 = 15

Answer 7

OMFG, LMFAO, thank you a billion tons..wow...it must've took hoooooourrrrssss....
but i very thank you as shown, and as said, for i also gave you medals and became your fan hahah
and i still THANK YOU VERY BERRY MUCH!!!!!!!11
;)

Answer 8

Question 1: Answer is \[8!/(8-6)! = 20160\] if order of doing chores matters; \[8!/(8!)(8-6)! = 28\] if it doesn't.

Question 2: Answer is \[6!/(6-2)!=30\] if order of delivery matters; \[6!/(2!)(6-2)!=15\] if it doesn't.

Answer 9

thank you @rchstudy,also!!!!!!
thank you!!!!!
<333