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combinatorics

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by jayhawk2001 » Sat Dec 08, 2007 9:26 am
Please post the source and the answer choices.

I'd go for 10 i.e. 4C3 + 3C1 + 3C1

where 4C3 is the number of ways of getting a 3 flower garland
from 4 different colors. 3C1 accounts for the extra blue and red
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by mleviko » Sat Dec 08, 2007 9:53 am
This was taken from the TestMagic forum, with no source or answer choices. The forum members said that the answer is 4, and think it's 10, as shown by the following non-mathematical solution:


we have six different flowers:

BLUE1
BLUE2
RED1
RED2
YELLOW
PINK

The requirements are :
1. make a 3-flower garland
2. the 3 flowers must be of different colors

All the possible ways of making 3 flower garlands with different colors are:

B1 R1 Y
B1 R1 P
B1 R2 Y
B1 R2 P
B1 Y P

B2 R1 Y
B2 R2 P
B2 Y P

R1 P Y

R2 P Y


so the answer is 10
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by swdatta » Tue Dec 11, 2007 10:10 pm
I have a question. Why are the combinations (B2 R2 Y) and (B2 R2 P) not included in the above mentioned list?

Thanks
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by gmatguy16 » Thu Dec 13, 2007 7:53 pm
i think answer is 12... possible combinations are:

bry or byp or brp or ryp
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by samirpandeyit62 » Fri Dec 14, 2007 12:13 am
IMO 10

2 blue
2 red
1 yellow
1 pink

coz each of the 2 blue colour flowers & 2 red will make the same colour combination

so basically we will have 2 possiblities

1) all 3 flowers are of diff colours i.e. 4C3 = 4ways

2) 2 flowers of the same colour & third of a diff colour

i.e 3 for blue & 3 for red

so total = 10 ways
Regards
Samir
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by gmatnew » Thu Jan 10, 2008 5:24 pm
i agree with gmatguy and swdatta,can some explain why 2 combinations mentioned by swdatta are not included even though other combinations with give the same color pattern are.
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by Stuart@KaplanGMAT » Thu Jan 10, 2008 6:17 pm
The question is a bit ambiguous.

If you don't need 3 different colours, then there are 10 possible garlands:

BBR
BBY
BBP
RRB
RRY
RRP
BRY
BRP
BYP
RYP

On the other hand, if you need 3 different colours, then there are only 4 possibilities:

BRY
BRP
BYP
RYP

Sometimes brute force (i.e. listing the combos) is much quicker than using formulas!
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by -Veronika- » Tue Feb 05, 2008 6:07 am
Please explain me why can't we use combinations formula here? choosing 3 out of 6? Thanks in advance!
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Its COMBInATION and not PERMUTATION

by deltasquare » Fri Feb 22, 2008 4:22 am
The answer, as TMers say, is 4.

Firstly, out of 6 flowers 3 can be chosen in 6C3 = 6 ways only. The order of choosing isn't important, what is important is the number of colours.. BRY is the same as RYB.. no need of counting this twice. Further this 6 ways has garlands that has 2 blue flowers, 2 red flowers as well. We have to remove these..

Since only 6 ways of choosing is possible, its better to proceed ahead by actually listing the ways.

BRY, BYP, BRP, PYR are the 4 ways in which colors are all different.. Hence answer is 4.
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Re: Its COMBInATION and not PERMUTATION

by Stuart@KaplanGMAT » Fri Feb 22, 2008 2:23 pm
deltasquare wrote:The answer, as TMers say, is 4.

Firstly, out of 6 flowers 3 can be chosen in 6C3 = 6 ways only. The order of choosing isn't important, what is important is the number of colours.. BRY is the same as RYB.. no need of counting this twice. Further this 6 ways has garlands that has 2 blue flowers, 2 red flowers as well. We have to remove these..

Since only 6 ways of choosing is possible, its better to proceed ahead by actually listing the ways.

BRY, BYP, BRP, PYR are the 4 ways in which colors are all different.. Hence answer is 4.
The ambiguity in the question is that "of different colors " doesn't specify whether all THREE flowers need to be different.
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