3 combination and permutation problems. Why combination here but permutation there?
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My understanding is that some problems can be solved both in ordered way and unordered way while some are more difficult and you have to choose one.
There are two problems that seem very similiar to me but one uses a permutation while other uses a combination. I would like to know why.
1. Classic birthday problem. As you know, $23$ people is what it takes to have the probability of a same bithday to be greater than $50%$. It is calculated this way.
Answer: $1 - frac{textrm{permut}(365, 23) }{ 365^{23}}$.
2. Coin flip problem. Flip a coin 100 times. $P(textrm{50 heads})$?
Answer: $frac{textrm{combin}(100, 50) }{ 2^{100}}$.
Why is permutation the correct formula for the first problem while combination is correct for the second problem?
Here's another problem. This one uses a combination with a multinomial coefficient.
3. Roll 6 dice. $P(textrm{3 pair})$? Example of a 3 pair would be, $(1, 1), (2, 2), (3, 3)$.
Answer: $frac{textrm{[combin}(6, 3) cdot 6!/(2!2!2!)]}{ 6^6}$
I think combining the combination with the multinomial coefficient, it orders the outcomes to some degree. Why do you have to order the outcomes in this problem while you don't in problem #2?
Thanks in advance for your help.
probability combinatorics permutations combinations
edited Nov 17 at 18:38
callculus
17.6k31427
asked Nov 17 at 18:22
user6877248
1
add a comment |
up vote
0
down vote
favorite
My understanding is that some problems can be solved both in ordered way and unordered way while some are more difficult and you have to choose one.
There are two problems that seem very similiar to me but one uses a permutation while other uses a combination. I would like to know why.
1. Classic birthday problem. As you know, $23$ people is what it takes to have the probability of a same bithday to be greater than $50%$. It is calculated this way.
Answer: $1 - frac{textrm{permut}(365, 23) }{ 365^{23}}$.
2. Coin flip problem. Flip a coin 100 times. $P(textrm{50 heads})$?
Answer: $frac{textrm{combin}(100, 50) }{ 2^{100}}$.
Why is permutation the correct formula for the first problem while combination is correct for the second problem?
Here's another problem. This one uses a combination with a multinomial coefficient.
3. Roll 6 dice. $P(textrm{3 pair})$? Example of a 3 pair would be, $(1, 1), (2, 2), (3, 3)$.
Answer: $frac{textrm{[combin}(6, 3) cdot 6!/(2!2!2!)]}{ 6^6}$
I think combining the combination with the multinomial coefficient, it orders the outcomes to some degree. Why do you have to order the outcomes in this problem while you don't in problem #2?
Thanks in advance for your help.
probability combinatorics permutations combinations
edited Nov 17 at 18:38
callculus
17.6k31427
asked Nov 17 at 18:22
user6877248
1
we use combinations when order does not matter. I.e. we toss a coin two times and we want to know probability of getting one tail. It does not matter whether we get tail first or second. On the other hand, if you have two people and you want to arrange them in line, (1,2) and (2,1) are two different arrangements.
– Vasya
Nov 17 at 18:31
@Vasya. You're saying question number 2, order doesn't matter so you use combination. So why use permutation in question number 1? We're not interested in arranging people in a line. We just want to know the probability of someone having the same birthday. In question number 3, why does the order matter when it doesn't matter in question number 2? Why isn't the answer to number 3 just combin(6, 3) / 6^6?
– user6877248
Nov 17 at 18:43
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My understanding is that some problems can be solved both in ordered way and unordered way while some are more difficult and you have to choose one.
There are two problems that seem very similiar to me but one uses a permutation while other uses a combination. I would like to know why.
1. Classic birthday problem. As you know, $23$ people is what it takes to have the probability of a same bithday to be greater than $50%$. It is calculated this way.
Answer: $1 - frac{textrm{permut}(365, 23) }{ 365^{23}}$.
2. Coin flip problem. Flip a coin 100 times. $P(textrm{50 heads})$?
Answer: $frac{textrm{combin}(100, 50) }{ 2^{100}}$.
Why is permutation the correct formula for the first problem while combination is correct for the second problem?
Here's another problem. This one uses a combination with a multinomial coefficient.
3. Roll 6 dice. $P(textrm{3 pair})$? Example of a 3 pair would be, $(1, 1), (2, 2), (3, 3)$.
Answer: $frac{textrm{[combin}(6, 3) cdot 6!/(2!2!2!)]}{ 6^6}$
I think combining the combination with the multinomial coefficient, it orders the outcomes to some degree. Why do you have to order the outcomes in this problem while you don't in problem #2?
Thanks in advance for your help.
probability combinatorics permutations combinations
edited Nov 17 at 18:38
callculus
17.6k31427
asked Nov 17 at 18:22
user6877248
1
My understanding is that some problems can be solved both in ordered way and unordered way while some are more difficult and you have to choose one.
There are two problems that seem very similiar to me but one uses a permutation while other uses a combination. I would like to know why.
1. Classic birthday problem. As you know, $23$ people is what it takes to have the probability of a same bithday to be greater than $50%$. It is calculated this way.
Answer: $1 - frac{textrm{permut}(365, 23) }{ 365^{23}}$.
2. Coin flip problem. Flip a coin 100 times. $P(textrm{50 heads})$?
Answer: $frac{textrm{combin}(100, 50) }{ 2^{100}}$.
Why is permutation the correct formula for the first problem while combination is correct for the second problem?
Here's another problem. This one uses a combination with a multinomial coefficient.
3. Roll 6 dice. $P(textrm{3 pair})$? Example of a 3 pair would be, $(1, 1), (2, 2), (3, 3)$.
Answer: $frac{textrm{[combin}(6, 3) cdot 6!/(2!2!2!)]}{ 6^6}$
I think combining the combination with the multinomial coefficient, it orders the outcomes to some degree. Why do you have to order the outcomes in this problem while you don't in problem #2?
Thanks in advance for your help.
probability combinatorics permutations combinations
probability combinatorics permutations combinations
edited Nov 17 at 18:38
callculus
17.6k31427
asked Nov 17 at 18:22
user6877248
1
edited Nov 17 at 18:38
callculus
17.6k31427
asked Nov 17 at 18:22
user6877248
1
edited Nov 17 at 18:38
callculus
17.6k31427
edited Nov 17 at 18:38
callculus
17.6k31427
edited Nov 17 at 18:38
callculus
17.6k31427
17.6k31427
asked Nov 17 at 18:22
user6877248
1
asked Nov 17 at 18:22
user6877248
1
asked Nov 17 at 18:22
user6877248
1
1
we use combinations when order does not matter. I.e. we toss a coin two times and we want to know probability of getting one tail. It does not matter whether we get tail first or second. On the other hand, if you have two people and you want to arrange them in line, (1,2) and (2,1) are two different arrangements.
– Vasya
Nov 17 at 18:31
@Vasya. You're saying question number 2, order doesn't matter so you use combination. So why use permutation in question number 1? We're not interested in arranging people in a line. We just want to know the probability of someone having the same birthday. In question number 3, why does the order matter when it doesn't matter in question number 2? Why isn't the answer to number 3 just combin(6, 3) / 6^6?
– user6877248
Nov 17 at 18:43
add a comment |
we use combinations when order does not matter. I.e. we toss a coin two times and we want to know probability of getting one tail. It does not matter whether we get tail first or second. On the other hand, if you have two people and you want to arrange them in line, (1,2) and (2,1) are two different arrangements.
– Vasya
Nov 17 at 18:31
@Vasya. You're saying question number 2, order doesn't matter so you use combination. So why use permutation in question number 1? We're not interested in arranging people in a line. We just want to know the probability of someone having the same birthday. In question number 3, why does the order matter when it doesn't matter in question number 2? Why isn't the answer to number 3 just combin(6, 3) / 6^6?
– user6877248
Nov 17 at 18:43
we use combinations when order does not matter. I.e. we toss a coin two times and we want to know probability of getting one tail. It does not matter whether we get tail first or second. On the other hand, if you have two people and you want to arrange them in line, (1,2) and (2,1) are two different arrangements.
– Vasya
Nov 17 at 18:31
we use combinations when order does not matter. I.e. we toss a coin two times and we want to know probability of getting one tail. It does not matter whether we get tail first or second. On the other hand, if you have two people and you want to arrange them in line, (1,2) and (2,1) are two different arrangements.
– Vasya
Nov 17 at 18:31
@Vasya. You're saying question number 2, order doesn't matter so you use combination. So why use permutation in question number 1? We're not interested in arranging people in a line. We just want to know the probability of someone having the same birthday. In question number 3, why does the order matter when it doesn't matter in question number 2? Why isn't the answer to number 3 just combin(6, 3) / 6^6?
– user6877248
Nov 17 at 18:43
@Vasya. You're saying question number 2, order doesn't matter so you use combination. So why use permutation in question number 1? We're not interested in arranging people in a line. We just want to know the probability of someone having the same birthday. In question number 3, why does the order matter when it doesn't matter in question number 2? Why isn't the answer to number 3 just combin(6, 3) / 6^6?
– user6877248
Nov 17 at 18:43
add a comment |
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we use combinations when order does not matter. I.e. we toss a coin two times and we want to know probability of getting one tail. It does not matter whether we get tail first or second. On the other hand, if you have two people and you want to arrange them in line, (1,2) and (2,1) are two different arrangements.
– Vasya
Nov 17 at 18:31
@Vasya. You're saying question number 2, order doesn't matter so you use combination. So why use permutation in question number 1? We're not interested in arranging people in a line. We just want to know the probability of someone having the same birthday. In question number 3, why does the order matter when it doesn't matter in question number 2? Why isn't the answer to number 3 just combin(6, 3) / 6^6?
– user6877248
Nov 17 at 18:43