Permutation and combination

[anteater85]

at a certain pet show, exactly 6 dogs are entered in the beagle division and exactly 6 are entered in the dalmation division. if the top 3 dogs in each division receive 1st,2nd and 3rd ribbons respectively, with no other dogs receiving a prize, how many different rosters of winners are possible for the two divisions together.

the answer is 14400

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[jay47]

at a certain pet show, exactly 6 dogs are entered in the beagle division and exactly 6 are entered in the dalmation division. if the top 3 dogs in each division receive 1st,2nd and 3rd ribbons respectively, with no other dogs receiving a prize, how many different rosters of winners are possible for the two divisions together.

the answer is 14400

Memorize the formulas for permutations and combinations:

Permutation (where order matters, as in this case): n! / (n-R)!

Combination (where order does not matter, as in just choosing three dogs, where there is no "placing")= n! / [R!(n-R)!]

Where "n" is the number you choose from (here 6 dogs) and "R" is the number you actually choose (3 dogs, order DOES matter, 1st, 2nd, and 3rd place)

So, do the math:

For each division: 6! / (6-3)!=

(6x5X4X3X2X1) / (3x2X1) = (notice 3x2X1 cancels out and you get....)

(6x5X4)=120

Now, you have two divisions, so for each of those divisions, you have 120 possibilities, meaning that each of those increase the number of different possibilities by 120, so the final answer is:

120 X 120 = 14,400

Easy as pie if you know the formula.


You can remember which one has the R! on the bottom because that one should have a smaller number of possibilities, combinations, because order does not matter.In this case order matters, because if dog "A" gets first place, that is not the same as dog "A" getting second, third and so forth!

Beatcha Street!

 

[whawha]

Are there any web source (or anything like that) where we could find a list of formulas and practice problems? Or maybe review videos? I have been trying to find one but it was not successful...

 

[razblo]

Memorize the formulas for permutations and combinations:

Permutation (where order matters, as in this case): n! / (n-R)!

Combination (where order does not matter, as in just choosing three dogs, where there is no "placing")= n! / [R!(n-R)!]

Where "n" is the number you choose from (here 6 dogs) and "R" is the number you actually choose (3 dogs, order DOES matter, 1st, 2nd, and 3rd place)

So, do the math:

For each division: 6! / (6-3)!=

(6x5X4X3X2X1) / (3x2X1) = (notice 3x2X1 cancels out and you get....)

(6x5X4)=120

Now, you have two divisions, so for each of those divisions, you have 120 possibilities, meaning that each of those increase the number of different possibilities by 120, so the final answer is:

120 X 120 = 14,400

Easy as pie if you know the formula.


You can remember which one has the R! on the bottom because that one should have a smaller number of possibilities, combinations, because order does not matter.In this case order matters, because if dog "A" gets first place, that is not the same as dog "A" getting second, third and so forth!

Beatcha Street!

That kind of helps. Thanks!

I don't ever remember seeing those formulas in pre-calculus, although I think the concept is somewhat common sense, so when that stuff showed up on some practice tests it kind of caught me off guard.

 

[anteater85]

That kind of helps. Thanks!

I don't ever remember seeing those formulas in pre-calculus, although I think the concept is somewhat common sense, so when that stuff showed up on some practice tests it kind of caught me off guard.

same here....I guess the only way u can learn is try to do the problem and then learn from it.(learn from ur mistakes)

 

[anteater85]

Country A and country B will form a joint committee on economic policy. The committee is to have exactly 6 members. 5 candidates for the committee come from Country A and 6 from Country B. if at least 3 members of the committee must come from Coutry A, how many distict committees are possible.

answer is 281

 

[razblo]

same here....I guess the only way u can learn is try to do the problem and then learn from it.(learn from ur mistakes)

I've gone through 2 Achiever QR tests and have scored 17 on both of them. I'm not really sure if that's good or bad, but I like to go through the test a second time and see which ones I got wrong and which ones I got right while re-doing problems either to increase speed or learn them.

 

[Streetwolf]

Country A and country B will form a joint committee on economic policy. The committee is to have exactly 6 members. 5 candidates for the committee come from Country A and 6 from Country B. if at least 3 members of the committee must come from Coutry A, how many distict committees are possible.

answer is 281

Try to solve yourself (answer below).

If at least 3 members must come from A then you are limited to only so many possibilities. For instance, the committee could NOT be formed with 2 members from A and 4 members from B. First write down your possibilities (remember that A has only 5 candidates). Then for each possibility, use a combination for both A and B to figure out how many ways they could select candidates under that possibility. Then add them all up at the end.










Hint: One possibility is 4 from A, 2 from B. For this case, use (5 C 4) to select the 4 representatives from A. Use (6 C 2) to select the 2 representatives from B. You get 5 and 15. Multiply them together to get 75. There are 75 ways to select 4 from A and 2 from B.























SOLUTION:

Possible choices:

3 from A, 3 from B = (5 C 3) * (6 C 3) = 10 * 20 = 200
4 from A, 2 from B = (5 C 4) * (6 C 2) = 5 * 15 = 75
5 from A, 1 from B = (5 C 5) * (6 C 1) = 1 * 6 = 6
6 from A, 0 from B CANNOT happen because A only has 5 to choose from.

Total is 200 + 75 + 6 = 281.

 

[predental1187]

Streetwolf, thanks for posting! that was a helpful explanation...I always seems to have trouble with these types of problems as well. If anyone has any more examples, please post!

 

[anteater85]

Streetwolf, thanks for posting! that was a helpful explanation...I always seems to have trouble with these types of problems as well. If anyone has any more examples, please post!

I've a lot of questions...i'll post them after my break...let's use thread to post up probability, combination and more word problems....

 

[jay47]

Try to solve yourself (answer below).

If at least 3 members must come from A then you are limited to only so many possibilities. For instance, the committee could NOT be formed with 2 members from A and 4 members from B. First write down your possibilities (remember that A has only 5 candidates). Then for each possibility, use a combination for both A and B to figure out how many ways they could select candidates under that possibility. Then add them all up at the end.

Hint: One possibility is 4 from A, 2 from B. For this case, use (5 C 4) to select the 4 representatives from A. Use (6 C 2) to select the 2 representatives from B. You get 5 and 15. Multiply them together to get 75. There are 75 ways to select 4 from A and 2 from B.


SOLUTION:

Possible choices:

3 from A, 3 from B = (5 C 3) * (6 C 3) = 10 * 20 = 200
4 from A, 2 from B = (5 C 4) * (6 C 2) = 5 * 15 = 75
5 from A, 1 from B = (5 C 5) * (6 C 1) = 1 * 6 = 6
6 from A, 0 from B CANNOT happen because A only has 5 to choose from.

Total is 200 + 75 + 6 = 281.

Good one Street. They probably won't get much harder than this one, so if you can understand this, then I think you can pretty much get all of them.
That is really a great question.....

 

[anteater85]

I was doing some question and came up with this question and thought it would be cool to tell u guys about it.

A computer program randomly creates triangle s w/ sides of integer lengths. If first two sides for a certain triangle are 3 and 8, what's the probability that the primeter of the triangle will be less than 20.


solution:
so as we know the side of the unknown should be less than 11 and bigger than five according to this rule:
3 + 8 =11
8-3=5

5<x<11

that would leave X to be 6.7.8.9.10

since we want the preimeter to be less than 20 and the sum of two known sides is 11, we want the unknown side to be less than 9.
so the desired outcome are 6,7,8

according to the probability formula=desired outcome/possible outcome=3/5

 

[jay47]

I was doing some question and came up with this question and thought it would be cool to tell u guys about it.

A computer program randomly creates triangle s w/ sides of integer lengths. If first two sides for a certain triangle are 3 and 8, what's the probability that the primeter of the triangle will be less than 20.


solution:
so as we know the side of the unknown should be less than 11 and bigger than five according to this rule:
3 + 8 =11
8-3=5

5<x<11

that would leave X to be 6.7.8.9.10

since we want the preimeter to be less than 20 and the sum of two known sides is 11, we want the unknown side to be less than 9.
so the desired outcome are 6,7,8

according to the probability formula=desired outcome/possible outcome=3/5

This is also a great question! Good one...

 

[anteater85]

A certain jewelry store sells customized rings in which three gemstones selected by the customer are set in a straight row along the band of the ring. if exactly 5 different types of gemstone are available, and if at least two of the gemstones in any given ring must be different, how many different rings are possible?

the answer is 120


here is how i solved it, can someone check if my solution is correct or I just got lucky to solve this question

I said since we want 3 gemstones for the rings we can have 2 possible way.
1st possibility is if all the stone are different so : 5!/3! =20
2nd possibility is when two of the stone must be different 5!/2!=60

20 x 60 =120

 

[anteater85]

A certain travel agency organizes sightseeing tours. Each client may select the cities she wishes to visit and the order in which she visits them. If each tour must include exactly 4 cities, and if there are exactly 10 cities available, how many different tours are possible?

here is how I solved but the answer is 5040

10!/4!(6!) = 210

 

[predental1187]

Is the rule for this that if you know 2 sides of a triangle, the third side cannot be smaller than their difference, or larger than their sum??

I think I vaguely remember this rule, but I've forgotten a lot of math concepts, as I haven't taken it in years!

I was doing some question and came up with this question and thought it would be cool to tell u guys about it.

A computer program randomly creates triangle s w/ sides of integer lengths. If first two sides for a certain triangle are 3 and 8, what's the probability that the primeter of the triangle will be less than 20.


solution:
so as we know the side of the unknown should be less than 11 and bigger than five according to this rule:
3 + 8 =11
8-3=5

5<x<11

that would leave X to be 6.7.8.9.10

since we want the preimeter to be less than 20 and the sum of two known sides is 11, we want the unknown side to be less than 9.
so the desired outcome are 6,7,8

according to the probability formula=desired outcome/possible outcome=3/5

 

[anteater85]

Is the rule for this that if you know 2 sides of a triangle, the third side cannot be smaller than their difference, or larger than their sum??

I think I vaguely remember this rule, but I've forgotten a lot of math concepts, as I haven't taken it in years!

yes that's right

 

[anteater85]

A child has 6 magnetic letters, C,H, I,I,P, P. how many different words can be spell by using these letters?

solution:
6!/ 2! 2! = 180

 

[anteater85]

How do you solve this?

A certain company sells jellybeans in the following 6 flavors only: banana, chocolate, grape, lemon, peach nad strawberry. If the jellybeans are sorted randomly into boxes containing exactly 2, 3 or 4 different flavors only, what is the probability that any given box contains grape jelly beans??

the answer is 1/2

 

[anteater85]

A promoter is organizing an opera concert. Ten singers are available for the program: 3 sopranos, 3 mezzos, 2 tenors and 2 baritones. if the promoter intends every segment to be duet by performers with different ocal ranges, how many pairings of performers are possible?

the answer is 37

 

[razblo]

I found this website...

http://sophistpundit.blogspot.com/2006/02/permutations-and-combinations.html

...and I think it does a really good job of explaining combination and permutation.

 

[Streetwolf]

How do you solve this?

A certain company sells jellybeans in the following 6 flavors only: banana, chocolate, grape, lemon, peach nad strawberry. If the jellybeans are sorted randomly into boxes containing exactly 2, 3 or 4 different flavors only, what is the probability that any given box contains grape jelly beans??

the answer is 1/2

1/3 chance of 2 flavor box
1/3 chance of 3 flavor box
1/3 chance of 4 flavor box

In the 2 flavor box, 1/3 chance of grape.
In the 3 flavor box, 1/2 chance of grape.
In the 4 flavor box, 2/3 chance of grape.

So answer is (1/3)(1/3) + (1/3)(1/2) + (1/3)(2/3) = (1/9) + (1/6) + (2/9) = (3/9) + (1/6) = 1/2

Answer 1/2.

 

[Streetwolf]

A promoter is organizing an opera concert. Ten singers are available for the program: 3 sopranos, 3 mezzos, 2 tenors and 2 baritones. if the promoter intends every segment to be duet by performers with different ocal ranges, how many pairings of performers are possible?

the answer is 37

3 soprano * 3 mezzo + 3 soprano * 2 tenor + 3 soprano * 2 baritone + 3 mezzo * 2 tenor + 3 mezzo * 2 baritone + 2 tenor * 2 baritone

9 + 6 + 6 + 6 + 6 + 4 = 37

 

[Streetwolf]

A child has 6 magnetic letters, C,H, I,I,P, P. how many different words can be spell by using these letters?

solution:
6!/ 2! 2! = 180

There are 6! ways to choose the order of the 6 pieces. But the Is and Ps repeat and having CHI(one)I(two)PP is no different than having CHI(two)I(one)PP and the same for the Ps. So in the denominator you get rid of the # of ways you can order each I and the # of ways you can order each P.

In this case, 2! for each of them.

6! / 2!2!

If you had CHIIPPPSSSS then there are 11 letters = 11!

In the denominator there are 2! ways to order the Is, there are 3! ways to order the Ps, and there are 4! ways to order the Ss, so you'd have

11! / 2!3!4! as your answer.

 

[Streetwolf]

A certain travel agency organizes sightseeing tours. Each client may select the cities she wishes to visit and the order in which she visits them. If each tour must include exactly 4 cities, and if there are exactly 10 cities available, how many different tours are possible?

here is how I solved but the answer is 5040

10!/4!(6!) = 210

ORDER MATTERS!!!!! It says so in the problem!!! Use a permutation.

 

[Streetwolf]

A certain jewelry store sells customized rings in which three gemstones selected by the customer are set in a straight row along the band of the ring. if exactly 5 different types of gemstone are available, and if at least two of the gemstones in any given ring must be different, how many different rings are possible?

the answer is 120


here is how i solved it, can someone check if my solution is correct or I just got lucky to solve this question

I said since we want 3 gemstones for the rings we can have 2 possible way.
1st possibility is if all the stone are different so : 5!/3! =20
2nd possibility is when two of the stone must be different 5!/2!=60

20 x 60 =120

2 of them must be different, AT LEAST. So all 3 could be different. Two possibilities.

#1 - You have 2 of one type, 1 of the other. Notice that it matters how you select them here!!! Why? Because one of them is used twice. So gemstone A and gemstone B is NOT the same as gemstone B and gemstone A (because in the first case you'll use A twice and in the second case you'll use B twice).

Permutation = (5 P 2) = 20

But now note that it matters how you place them on the ring! You can have the lone gemstone in 3 spots. So a total of 60 ways to do it.

#2 - you have 3 different types. But it still matters how you pick them because it matters which order they go on the ring. You have a permutation:

(5 P 3) = 60

Total of 60 + 60 = 120

 

[kkmaths90]

1/3 chance of 2 flavor box
1/3 chance of 3 flavor box
1/3 chance of 4 flavor box

In the 2 flavor box, 1/3 chance of grape.
In the 3 flavor box, 1/2 chance of grape.
In the 4 flavor box, 2/3 chance of grape.

So answer is (1/3)(1/3) + (1/3)(1/2) + (1/3)(2/3) = (1/9) + (1/6) + (2/9) = (3/9) + (1/6) = 1/2

Answer 1/2.

Thanks for all explanations...i got all of them except this one...I got secon part of 1/3 , 1/2 and 2/3 chances

but not the part where it says:
1/3 change of 2 flavor box, 1/3 chance of 3 flavor box and 1/3 chance of 4 flavor box ...please explain that part. Thank you

 

[Streetwolf]

Thanks for all explanations...i got all of them except this one...I got secon part of 1/3 , 1/2 and 2/3 chances

but not the part where it says:
1/3 change of 2 flavor box, 1/3 chance of 3 flavor box and 1/3 chance of 4 flavor box ...please explain that part. Thank you

They are randomly sorted into boxes of 2, 3, and 4. So there is a 1/3 chance of being sorted into any one of those.