Question

I have 5 pieces of paper, each with a different math problem. In how many ways can he give these problems to his 10 friends?

Answer 1

hey @JuanSalitras

Answer 2

im not great with permutations or combinations... but I think i can do this one, lemme know if it seems wrong tho and i''ll ask someone else to step in who knows more than i do about them

Answer 3

now, i'm pretty sure this one falls under "permutations":
as we can give more than one question to one person, it should be fairly simple:
so there are 10 possible choices
and 5 questions
so 10 x 10 x 10 x 10 x 10
n^r = 10^5 = 100000 ... yeah?

Answer 4

Thank you so much for your help.

Answer 5

ur right dude, any time

Answer 6

Why would the answer not be 10*9*8*7*6 ?

I am not saying that 10^5 is incorrect but wandering why
10*9*8*7*6 = 30,240 would not be correct.

5 people and 10 problems: first person has 10 problem choices, second has 9, third has 8, fourth has 7, and third has 6.

Answer 7

the OP has edited the question, initially he stated that more than 1 question could go to the same person, therefore the choices are not reduced with each new question
ie exponential solve rather than factorial @Directrix

Answer 8

also, u have it backwards, 10 people and 5 problems, not 5 people and 10 problems

Answer 9

>>the OP has edited the question -- that explains my confusion. Thanks.