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Maths GCSE 9-1 Product Rules Question
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There are 52 cards in a deck. Tom is going to give two cards to Jay. How many different pairs of cards could Jay get?
For this question the answer was (52 x 51)/2. I don't understand why the answer was divided by 2?
There are 52 cards in a deck. Peter is going to give one to Casper and one to Kelly. How many different ways of doing this are there?
The answer to this was 52x51. As someone else on the TSR pointed out, wouldn't you multiply the answer by 2?
For this question the answer was (52 x 51)/2. I don't understand why the answer was divided by 2?
There are 52 cards in a deck. Peter is going to give one to Casper and one to Kelly. How many different ways of doing this are there?
The answer to this was 52x51. As someone else on the TSR pointed out, wouldn't you multiply the answer by 2?
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#2
52 options for first card. 51 for second.
52 x 51 are the number of pairs.
However, note the questions says "different".
Observe that A, B = B,A. So 52x51 will have these twice over hence divide by 2 (as you over counted).
52 x 51 are the number of pairs.
However, note the questions says "different".
Observe that A, B = B,A. So 52x51 will have these twice over hence divide by 2 (as you over counted).
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#4
Okay so the first one you divide by two because for every two cards, there is one way of giving it. Let's call the first card A and the second B. Tom can give Jay the combination 'A and B'. He can also give the combination 'B and A'. However, these are the same combination of cards. This is true for every pair of cards you get, so you need to divide the whole value by two.
For the second one, however, 'A and B' is different to 'B and A' because there are two different people holding the cards. It is not a combination but rather two separate people taking cards at once. What I mean by this is that Casper can get A and Kelly can get B, but this is different to Casper getting B and Kelly getting A.
For the second one, however, 'A and B' is different to 'B and A' because there are two different people holding the cards. It is not a combination but rather two separate people taking cards at once. What I mean by this is that Casper can get A and Kelly can get B, but this is different to Casper getting B and Kelly getting A.
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(Original post by CapitalistAlgan)
Okay so the first one you divide by two because for every two cards, there is one way of giving it. Let's call the first card A and the second B. Tom can give Jay the combination 'A and B'. He can also give the combination 'B and A'. However, these are the same combination of cards. This is true for every pair of cards you get, so you need to divide the whole value by two.
For the second one, however, 'A and B' is different to 'B and A' because there are two different people holding the cards. It is not a combination but rather two separate people taking cards at once.
Okay so the first one you divide by two because for every two cards, there is one way of giving it. Let's call the first card A and the second B. Tom can give Jay the combination 'A and B'. He can also give the combination 'B and A'. However, these are the same combination of cards. This is true for every pair of cards you get, so you need to divide the whole value by two.
For the second one, however, 'A and B' is different to 'B and A' because there are two different people holding the cards. It is not a combination but rather two separate people taking cards at once.
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#7
Spoiler:
Also, once you get to higher Maths:
You learn about combinations and permtuations. Learn about these - it'll increase your understanding.
In this case, the amount of permutations for picking 2 cards from 52 is simply 52!/(52-2)! = 2652. But in this case, permutations count a different order as a different combination e.g. if you had A, B - A,B will count as a different permutation to B,A. So you have 2652 ways of picking 2 cards from 52 but notice these 2 cards come in pair and if you reverse the order, you get a different permutation (but not a unique combination or pair).
The amount of combinations however is 52!/(52-2)!2! giving you the correct answer of 1326 ( = 51 x 52 / 2).
Permuations of picking Y things from X objects is X!/(X-Y)!.
Combinations of picking Y things from X objects is X/(X-Y)!Y!. The difference in the formula is because you would have divided by Y! because you are eliminating the same order combinations; if you have Y cards, the amount of ways organising them is Y! and since combinations consider a rearrangement of the set the same combination, you would have to divide by Y!.
Show
Also, once you get to higher Maths:
You learn about combinations and permtuations. Learn about these - it'll increase your understanding.
In this case, the amount of permutations for picking 2 cards from 52 is simply 52!/(52-2)! = 2652. But in this case, permutations count a different order as a different combination e.g. if you had A, B - A,B will count as a different permutation to B,A. So you have 2652 ways of picking 2 cards from 52 but notice these 2 cards come in pair and if you reverse the order, you get a different permutation (but not a unique combination or pair).
The amount of combinations however is 52!/(52-2)!2! giving you the correct answer of 1326 ( = 51 x 52 / 2).
Permuations of picking Y things from X objects is X!/(X-Y)!.
Combinations of picking Y things from X objects is X/(X-Y)!Y!. The difference in the formula is because you would have divided by Y! because you are eliminating the same order combinations; if you have Y cards, the amount of ways organising them is Y! and since combinations consider a rearrangement of the set the same combination, you would have to divide by Y!.
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(Original post by thekidwhogames)
Spoiler:
Also, once you get to higher Maths:
You learn about combinations and permtuations. Learn about these - it'll increase your understanding.
In this case, the amount of permutations for picking 2 cards from 52 is simply 52!/(52-2)! = 2652. But in this case, permutations count a different order as a different combination e.g. if you had A, B - A,B will count as a different permutation to B,A. So you have 2652 ways of picking 2 cards from 52 but notice these 2 cards come in pair and if you reverse the order, you get a different permutation (but not a unique combination or pair).
The amount of combinations however is 52!/(52-2)!2! giving you the correct answer of 1326 ( = 51 x 52 / 2).
Permuations of picking Y things from X objects is X!/(X-Y)!.
Combinations of picking Y things from X objects is X/(X-Y)!Y!. The difference in the formula is because you would have divided by Y! because you are eliminating the same order combinations; if you have Y cards, the amount of ways organising them is Y! and since combinations consider a rearrangement of the set the same combination, you would have to divide by Y!.
Show
Also, once you get to higher Maths:
You learn about combinations and permtuations. Learn about these - it'll increase your understanding.
In this case, the amount of permutations for picking 2 cards from 52 is simply 52!/(52-2)! = 2652. But in this case, permutations count a different order as a different combination e.g. if you had A, B - A,B will count as a different permutation to B,A. So you have 2652 ways of picking 2 cards from 52 but notice these 2 cards come in pair and if you reverse the order, you get a different permutation (but not a unique combination or pair).
The amount of combinations however is 52!/(52-2)!2! giving you the correct answer of 1326 ( = 51 x 52 / 2).
Permuations of picking Y things from X objects is X!/(X-Y)!.
Combinations of picking Y things from X objects is X/(X-Y)!Y!. The difference in the formula is because you would have divided by Y! because you are eliminating the same order combinations; if you have Y cards, the amount of ways organising them is Y! and since combinations consider a rearrangement of the set the same combination, you would have to divide by Y!.
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#11
(Original post by 2018GCSEVictim)
Going to need some time to wrap my head around that. Thanks for the heads up as I'm planning to take Maths in the future.
Going to need some time to wrap my head around that. Thanks for the heads up as I'm planning to take Maths in the future.
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(Original post by thekidwhogames)
No problem - I agree combinatorics is quite a tough subject in Maths. You do cover a little of it at A level but to appreciate the beautify of it, I'd recommend reading some Olympiad books on it in the future. Good luck in the GCSEs (bloody hate the 9-1 system too).
No problem - I agree combinatorics is quite a tough subject in Maths. You do cover a little of it at A level but to appreciate the beautify of it, I'd recommend reading some Olympiad books on it in the future. Good luck in the GCSEs (bloody hate the 9-1 system too).
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