I am helping a friend learn math and sometimes he gives me problems I cannot figure out.
This is a discrete problem, permutations and combinations. Anyone able to help me?
Four people are to be selected at random from a group of 4 couples. In how many ways can this be done, given the following conditions:
a) No restrictions
b) The group must have at least one couple.
c) Each couple must be represented in the group.
The answer (according to the book) are 70, 54, 16 and I cannot get any of these, I hate to admit.
I’m going to beg your indulgence
7 hours ago
2 comments:
That's a fun problem. Thanks for sharing.
If there are no restrictions, you can think of 4 slots open and 8 people to choose from. There are 8 ways to fill the first slot, once that's chosen, there are 7 ways to fill the 2nd, then 6 ways to fill the 3rd, then 5 ways to fill the 4th. However, you've double counted this way, so you have to get rid of the doubles. For every one of your choices, say "a1,b1,c2,d1" (couples a,b,c,d and partner 1 or 2), there are 4*3*2*1 ways to have chosen that one in your above counting way: "b1,a1,c2,d1" is a sample duplicate. SO... your final answer to "no restrictions is: (8*7*6*5)/(4*3*2*1) = 70
For the 3rd one (every couple represented): in your 4 choices of people to pick, you can pick 2 from couple a, then 2 from couple b, then 2 from couple c, and 2 from couple d. So 2*2*2*2=16 ways.
For the 2nd on (at least one full couple):
"A" is the only couple then I counted 4 ways to pick one from b and one from c, 4 ways to pick one from b and one from d, and 4 ways to pick 1 from c and one from d. This gives 12.
"AB" 2 couples, "AC" 2 couples, "AD" 2 couples. Count so far is 15.
"B" is the only couple ... 12 as before. and "BC", "BD". Count so far is 29.
"C" is the only couple ... 12 as before. and "CD". Count so far is 42.
"D" is the only couple ... 12 as before. Count is 54.
Ms. Cookie
Thank you so much!! Great explanation!
I sent the answer to my friend.
I love your blog - I cannot remember if I link to it but I pick it up from Math Tales From the Spring. (I will add a link here)
I love tutoring my friend - it makes me jump around in different courses and work harder than my students are making me work. That (and blogs like yours) are helping e not rust.
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