anonymous
  • anonymous
MEDAL!! (suppose there is a group of 7 people from which we will make a committee) In how many ways can we pick a three-person committee?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jim_thompson5910
  • jim_thompson5910
Let's say we have 3 slots A,B,C how many do we have to choose from for slot A?
anonymous
  • anonymous
7?
mathmate
  • mathmate
Correct, then for slot B?

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mathmate
  • mathmate
Recall that one of the sever is already in slot A.
anonymous
  • anonymous
6?
jim_thompson5910
  • jim_thompson5910
yep, and then C?
anonymous
  • anonymous
5
anonymous
  • anonymous
multiply?
jim_thompson5910
  • jim_thompson5910
yes, you'll multiply 7, 6 and 5
anonymous
  • anonymous
210
jim_thompson5910
  • jim_thompson5910
that won't be your final answer but it gets you closer
jim_thompson5910
  • jim_thompson5910
210 is the number of ways to pick 3 people IF order mattered but there is no ranking on this committee, so order doesn't matter
jim_thompson5910
  • jim_thompson5910
so what you have to do is divide by 3! = 3*2*1 = 6 to get the correct count
jim_thompson5910
  • jim_thompson5910
the reason why is because there are 6 ways to order any three objects xyz xzy yxz yzx zxy zyx
anonymous
  • anonymous
210 divided by 3?
jim_thompson5910
  • jim_thompson5910
210/6 actually
jim_thompson5910
  • jim_thompson5910
or 210/(3!)
anonymous
  • anonymous
ah okay. I read your response wrong haha so 35
jim_thompson5910
  • jim_thompson5910
correct
jim_thompson5910
  • jim_thompson5910
another way is to use the combination formula \[\Large _n C _r = \frac{n!}{r!*(n-r)!}\] with n = 7 and r = 3 and you'll get the same answer
anonymous
  • anonymous
@jim_thompson5910 what if it was a four-person committee instead? wouldn't it be 7C4=35? also?
jim_thompson5910
  • jim_thompson5910
that is correct
anonymous
  • anonymous
thank you
jim_thompson5910
  • jim_thompson5910
if you think of it in terms of 7 slots A through G we have this |dw:1441932178631:dw|
jim_thompson5910
  • jim_thompson5910
now imagine cutting a line through that group such that one side (say the left side) has 3 slots and the other side has 4 slots |dw:1441932241059:dw|
jim_thompson5910
  • jim_thompson5910
arranging 7 people to go in slots A through C is the exact same as arranging the remaining 4 to go in D through G so that's why 7 C 3 = 7 C 4 in general \[\LARGE _n C _x = \ _n C _y\] where x+y = n (so in this case, x+y = 3+4 = 7 which is the value of n)
anonymous
  • anonymous
Yes, that is the way my instructor explains it also, thanks again!
jim_thompson5910
  • jim_thompson5910
you're welcome

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