Math question

[chordata]

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Is there an easy way to set up an equation for the following problem:
A florist has 3 pink, 2 yellow, and 5 red roses. How many three rose combinations are possible?

 

[Richnator]

I would use the combinations formula for that problem.

 

[joonkimdds]

Is there an easy way to set up an equation for the following problem:
A florist has 3 pink, 2 yellow, and 5 red roses. How many three rose combinations are possible?

let me guess...3 times 2 times 5? I don't know just came up on my head. is it right?

 

[onetoothleft]

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.

 

[chordata]

Thanks for your help.

I approached the problem the same way.

I was just hoping there would be an easier way to solve it.

 

[BenignDMD]

Is there an easy way to set up an equation for the following problem:
A florist has 3 pink, 2 yellow, and 5 red roses. How many three rose combinations are possible?

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

 

[chordata]

What is the combination formula?

Are you referring to 10!/6!4!? (Where order doesn't matter.)

 

[chordata]

Has anyone encountered any half angle identity questions on the DAT?

 

[Richnator]

Well, it is actually 10!/((10-3)!3!) if I am not mistaken

 

[mochafreak]

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]

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.

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.

 

[Richnator]

Umm.... Was that in english?

 

[joonkimdds]

There should be 9

3 pink 2 yellow 5 red

ppp
ppy
pyy
rrr
ryp
rry
rrp
ppr
yyr

so...we can have ppy but we can't have pyp?

 

[dat_student]

so...we can have ppy but we can't have pyp?

ppy = pyp if in circular arrangement

...p.......p......y
../.\..= /.\ =../.\ [clockwise rotation]
.p..y...y...p..p..p

 

[joonkimdds]

ppy = pyp if in circular arrangement

...p.......p......y
../.\..= /.\ =../.\ [clockwise rotation]
.p..y...y...p..p..p

how come Kaplan blue book doesn't have this kind of question?