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sasogeek
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6 friends went for a party at the house of one of their friends, so there was 7 of them at the party. during the party, they wanted to buy tickets for a lottery and the required number of numbers on the lottery tickets was 5 out of 90 possible numbers (190). if each of them was to write only one number and pass it on to another until there's 5 numbers on the tickets. how many tickets should they buy? and how many possible ways can they write 5 different numbers on each ticket?
 one year ago
 one year ago
sasogeek Group Title
6 friends went for a party at the house of one of their friends, so there was 7 of them at the party. during the party, they wanted to buy tickets for a lottery and the required number of numbers on the lottery tickets was 5 out of 90 possible numbers (190). if each of them was to write only one number and pass it on to another until there's 5 numbers on the tickets. how many tickets should they buy? and how many possible ways can they write 5 different numbers on each ticket?
 one year ago
 one year ago

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sasogeek Group TitleBest ResponseYou've already chosen the best response.0
source: http://www.cims.nyu.edu/~regev/teaching/discrete_math_fall_2005/dmbook.pdf i understand that the number of possible ways they can write 5 numbers = 90*89*87*86*85 according to the book, because some 5 sets of numbers may be repeated, they'd have to buy 90*89*88*87*86/5*4*3*2*1 what i'm not getting is, if there's 7 people at the party, why don't they buy \(\large 7*(\frac{90*89*88*87*86}{5*4*3*2*1}) \) number of tickets?
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
or even \(\large \frac{90*89*88*87*86}{7*6*5*4*3*2*1} \) ?
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
When you say "how many tickets should they buy" do you mean what is the minimum number of tickets they can buy whilst all having selected an even amount of numbers? As in first pass around7 numbers are selected, second pass14 numbers, 3rd21, 4th28, 5th35. 35 being the lowest number divisible by 7?
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
no, they want to buy tickets such that every possible combination of numbers that the winning ticket must have is written on one of the tickets they buy.
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
Oh I see! Idiots!
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
they may choose any random number but it may not be repeated by the next person. so if the first person chooses to write 50, the second person may choose to write 32, and same with the rest of the friends at the party.
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
they may be idiots, but their problem has to be solved :)
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
@satellite73 I could use your help :) very much needed xD
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
@TuringTest I need thy help asap :)
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
im pretty sure the answer is 54891018
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
So are they allowed to repeat?
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
To be honest, I'm really not the best person to be asking as I have not even started my maths degree yet. I don't want to give you any wrong answers. Hopefully these permutation calculators help: http://www.statisticshowto.com/calculators/permutationcalculatorandcombinationcalculator/ http://www.mathsisfun.com/combinatorics/combinationspermutationscalculator.html Good luck!
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Toughie. I'm not the best at these bro
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
Does this help? n=90 r=5 Number of possible Combinations = n! r!(n – r)! = 90! 5!(90 – 5)! = 43949268
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
Or \[43949268\times7=307644876\]
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
i'm not making any sense out of that, there's 7 people involved, do you consider how many of them that are there or it doesn't matter and all that matters is just the number of numbers?
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
I dont think the number of people there is relevant, therefore i would guess that the answer is 43949268. But honestly, im not too sure if that actually the answer you are after. What do you think?
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
your answer is the same provided in the book (link to source in my first post up above (page 8)), i'm just confused as to why the number of people wasn't taken into consideration :/
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
Ahhh, ok. If you think about it 1person can do the same job as the 7 people, all they are doing is picking numbers.
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
Similarly, if i was with you and we were trying to find out the probability of a certain number on a dice being rolled it would be irrelevant who was doing the rolling or if we took in turns. All that matters is the amount of times the dice is rolled, its relevant who is doing the rolling.
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
you mean it is irrelevant who is doing the rolling... right?
 one year ago

sasogeek Group TitleBest ResponseYou've already chosen the best response.0
ok thanks :) now i'm settled xD
 one year ago

nabizag Group TitleBest ResponseYou've already chosen the best response.1
Haha! I'm glad I could have been of some help!
 one year ago
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