sasogeek
  • sasogeek
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 (1-90). 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?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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sasogeek
  • sasogeek
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?
anonymous
  • anonymous
5 tickets
sasogeek
  • sasogeek
or even \(\large \frac{90*89*88*87*86}{7*6*5*4*3*2*1} \) ?

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More answers

anonymous
  • anonymous
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 around-7 numbers are selected, second pass-14 numbers, 3rd-21, 4th-28, 5th-35. 35 being the lowest number divisible by 7?
sasogeek
  • sasogeek
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.
anonymous
  • anonymous
Oh I see! Idiots!
sasogeek
  • sasogeek
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.
sasogeek
  • sasogeek
they may be idiots, but their problem has to be solved :)
sasogeek
  • sasogeek
@satellite73 I could use your help :) very much needed xD
sasogeek
  • sasogeek
@TuringTest I need thy help asap :)
anonymous
  • anonymous
im pretty sure the answer is 54891018
anonymous
  • anonymous
So are they allowed to repeat?
anonymous
  • anonymous
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/permutation-calculator-and-combination-calculator/ http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html Good luck!
Hero
  • Hero
Toughie. I'm not the best at these bro
anonymous
  • anonymous
Does this help? n=90 r=5 Number of possible Combinations = n! r!(n – r)! = 90! 5!(90 – 5)! = 43949268
anonymous
  • anonymous
Or \[43949268\times7=307644876\]
sasogeek
  • sasogeek
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?
anonymous
  • anonymous
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?
sasogeek
  • sasogeek
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 :/
anonymous
  • anonymous
Ahhh, ok. If you think about it 1person can do the same job as the 7 people, all they are doing is picking numbers.
anonymous
  • anonymous
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.
sasogeek
  • sasogeek
you mean it is irrelevant who is doing the rolling... right?
anonymous
  • anonymous
lol yep!
sasogeek
  • sasogeek
ok thanks :) now i'm settled xD
anonymous
  • anonymous
Haha! I'm glad I could have been of some help!

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