Math question
Started by chordata
chordata said:
let me guess...3 times 2 times 5? I don't know just came up on my head. is it right?
There should be 9
3 pink 2 yellow 5 red
ppp
ppy
pyy
rrr
ryp
rry
rrp
ppr
yyr
you can't use the typical combination formula here since you have 3 different colors of flowers . . . so I just worked it out logically since there should be relatively few combinations.
1. There is only one possibility that has all three colors: ryp
2. Red+Yellow: rry, ryy
3. Yellow+Pink: ppy, pyy
4. Pink+Red: rrp, rrp
5. Only Pink: ppp
6. Only Red: rrr
7. Only Yellow: not enough yellow flowers
that totals 9.
Remember that this question is asking for combinations, not probabilities. So the fact that there are so many red roses does not matter, you can only use up to 3 anyway.
3 pink 2 yellow 5 red
ppp
ppy
pyy
rrr
ryp
rry
rrp
ppr
yyr
you can't use the typical combination formula here since you have 3 different colors of flowers . . . so I just worked it out logically since there should be relatively few combinations.
1. There is only one possibility that has all three colors: ryp
2. Red+Yellow: rry, ryy
3. Yellow+Pink: ppy, pyy
4. Pink+Red: rrp, rrp
5. Only Pink: ppp
6. Only Red: rrr
7. Only Yellow: not enough yellow flowers
that totals 9.
Remember that this question is asking for combinations, not probabilities. So the fact that there are so many red roses does not matter, you can only use up to 3 anyway.
chordata said:
deoends what kind of combinations they want...did the problem specify if you could choose 3 reds as a combo, or 3 pinks, instead of pyr or yrr? what are the answer choices...b/c if you use the combinations formula for this, which it is applicable, you get 210 different combinations...look at the answer choices, that will tell you what formula they want you to use
The combinations (or binomial) formula does generalize to a multinomial, which counts the number of ways to obtain a specified combination. While that could be used here, you still have to enumerate each combination and you really don't gain anything. What is needed is the multichoose formula, which counts how many ways to form k-sets of n symbols (being n+k-1 choose k). Subtract 1 for the one impossible combination (yyy), and you have the answer.
mochafreak said:
If anyone remembers my posts referring to QR, I suck at QR. I read that question, picked 'B' and moved on. My husband who is a physics prof just got bored and decided to contribute to the forum. Hence, my first intelligent QR post. 😉 Now, when I get a 14 on QR tomorrow this post will make sense.
onetoothleft said:
so...we can have ppy but we can't have pyp?
joonkimdds said:
ppy = pyp if in circular arrangement
...p.......p......y
../.\..= /.\ =../.\ [clockwise rotation]
.p..y...y...p..p..p
dat_student said:
how come Kaplan blue book doesn't have this kind of question?
