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sasogeek
 3 years ago
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?
sasogeek
 3 years ago
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?

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sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0source: 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?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When 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?

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0no, 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.

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0they 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.

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0they may be idiots, but their problem has to be solved :)

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 I could use your help :) very much needed xD

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0@TuringTest I need thy help asap :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im pretty sure the answer is 54891018

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So are they allowed to repeat?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To 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!

Hero
 3 years ago
Best ResponseYou've already chosen the best response.0Toughie. I'm not the best at these bro

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does this help? n=90 r=5 Number of possible Combinations = n! r!(n – r)! = 90! 5!(90 – 5)! = 43949268

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Or \[43949268\times7=307644876\]

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0i'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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0your 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 :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ahhh, ok. If you think about it 1person can do the same job as the 7 people, all they are doing is picking numbers.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Similarly, 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.

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0you mean it is irrelevant who is doing the rolling... right?

sasogeek
 3 years ago
Best ResponseYou've already chosen the best response.0ok thanks :) now i'm settled xD

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Haha! I'm glad I could have been of some help!
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