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21 Jul 2016, 12:48
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Hi All,
I am struggling with Permutations and Combinations. I studied the Permutations and Combination post by Bunuel and others on GMAT. I also read some articles online. I am beginning to get an understanding of the two. But every once in a while I run into something that just defies what I THOUGHT was the right approach. Please help me confirm/correct my approach. To keep it simple where I can hopefully jot down various arrangements for a better visualization, I have picked reasonably small numbers.
Arrangements:
I have 3 marbles of different colors (Red, Blue, Yellow). In how many ways can I arrange these three marbles?
n=3
3! = 6
RBY
RYB
BRY
BYR
YRB
YBR
Combination:
I have 5 slots and 3 red marbles. In how many ways can I fill up those 5 slots?
n=5, r=3
5C2 = 10. Correct?
1. RRR__
2. RR_R_
3. RR__R
4. R_RR_
5. R__RR
6. _RRR_
7. __RRR
8. _RR_R
9. _R_RR
10. R_R_R
Permutations:
I have 5 slots and one Red, one Blue and one Yellow marbles. In how many ways can I fill up those 5 slots?
n=5, r=3
5P2 = 60
OR
5C2 * 3! because you can arrange two marbles in 10 combinations and for each combination you have six ways to order R, B and Yellow.
1.1 RBY__
1.2 RYB__
1.3 YRB__
1.4 YBR__
1.5 BYR__
1.6 BRY__
and so on from 2.1 - 10.6
Hopefully I am on the right track thus far in my logic.
But here is where I get tangled in my own thoughts.
Arrangements:
I have two Red, one Blue and one Yellow marbles. In how many ways can I arrange them?
Had it been all 4 different colors, I know I would go for n! because there are 4 ways to select the first, 3 ways for the second and so on. In this case, since two are Red, the order for those two doesn't matter. But the order of Blue and Yellow matters. So I approach them as an arrangement of 3 objects as in (RR) B and Y which gives me 6 arrangements where two Red marbles are always placed next to each other.
1. (RR) B Y
2. B (RR) Y
3. Y (RR) B
4. (RR) Y B
5. B Y (RR)
6. Y B (RR)
I also need to account for the possibilities where the Reds are separated.
1. R B R Y
2. R Y R B
3. B R Y R
4. Y R B R
5. R Y B R
6. R B Y R
How do I add these last 6 arrangements into the above grouping with a formula? Is there a better way to approach this kind of a problem?
Also, please provide an example of a very similar situation with a bit larger numbers (say (A) total arrangements for 4 reds, 3 yellows and 2 blue for a total of 9 marbles (B) combinations of 4 marbles such that there is at least one red and one blue).
Thanks.
I am struggling with Permutations and Combinations. I studied the Permutations and Combination post by Bunuel and others on GMAT. I also read some articles online. I am beginning to get an understanding of the two. But every once in a while I run into something that just defies what I THOUGHT was the right approach. Please help me confirm/correct my approach. To keep it simple where I can hopefully jot down various arrangements for a better visualization, I have picked reasonably small numbers.
Arrangements:
I have 3 marbles of different colors (Red, Blue, Yellow). In how many ways can I arrange these three marbles?
n=3
3! = 6
RBY
RYB
BRY
BYR
YRB
YBR
Combination:
I have 5 slots and 3 red marbles. In how many ways can I fill up those 5 slots?
n=5, r=3
5C2 = 10. Correct?
1. RRR__
2. RR_R_
3. RR__R
4. R_RR_
5. R__RR
6. _RRR_
7. __RRR
8. _RR_R
9. _R_RR
10. R_R_R
Permutations:
I have 5 slots and one Red, one Blue and one Yellow marbles. In how many ways can I fill up those 5 slots?
n=5, r=3
5P2 = 60
OR
5C2 * 3! because you can arrange two marbles in 10 combinations and for each combination you have six ways to order R, B and Yellow.
1.1 RBY__
1.2 RYB__
1.3 YRB__
1.4 YBR__
1.5 BYR__
1.6 BRY__
and so on from 2.1 - 10.6
Hopefully I am on the right track thus far in my logic.
But here is where I get tangled in my own thoughts.
Arrangements:
I have two Red, one Blue and one Yellow marbles. In how many ways can I arrange them?
Had it been all 4 different colors, I know I would go for n! because there are 4 ways to select the first, 3 ways for the second and so on. In this case, since two are Red, the order for those two doesn't matter. But the order of Blue and Yellow matters. So I approach them as an arrangement of 3 objects as in (RR) B and Y which gives me 6 arrangements where two Red marbles are always placed next to each other.
1. (RR) B Y
2. B (RR) Y
3. Y (RR) B
4. (RR) Y B
5. B Y (RR)
6. Y B (RR)
I also need to account for the possibilities where the Reds are separated.
1. R B R Y
2. R Y R B
3. B R Y R
4. Y R B R
5. R Y B R
6. R B Y R
How do I add these last 6 arrangements into the above grouping with a formula? Is there a better way to approach this kind of a problem?
Also, please provide an example of a very similar situation with a bit larger numbers (say (A) total arrangements for 4 reds, 3 yellows and 2 blue for a total of 9 marbles (B) combinations of 4 marbles such that there is at least one red and one blue).
Thanks.
21 Jul 2016, 13:35
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Witchcrafts
Always remember as a thumb rule below go through the question types:
For permutations, "Find the number of permutations tha can be made...." OR "Find the number of arrangements that can be made..." OR "Find the number of ways in which you can arrange...".
For combinations : "Find the number of combinations that can be made..." OR "Find the number of selections that can be made...." OR "Find the number of ways in which you can select".
The number of permutations of n distinct items taking r at a time is nPr = n! / (n-r)!.
The number of combinations of n dissimilar things taken r at a time is nCr = n! /r! (n-r)!.
21 Jul 2016, 15:53
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msk0657
I made a typo. r = 3 not 2. But I guess you made a mistake too?\(5C3 = 5!/(3!*2!)\) = 10 not 20, correct? I have made a list of all the possible arrangements above. I have come up with only 10. So just want to confirm I am not missing anything here.
