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09 Nov 2010, 09:55
Kudos
Bookmarks
Hi,
This questions is not specific but a general question about combinatorics, I wasn't sure where to post it, so I hope I posted it in the right forum.
I need some help understanding a Combinatorics concept. I have been using the anagram method and the combinations formula.
I want to understand how to calculate the maximum possible combinations (order does not matter) for questions like:
[Mr. and Mrs. X want to have 4 children...] - At this point how do I calculate all the possible combinations of Boy and Girls they can have if they have 4 children.
One way is the Slot method,
2 (B/G) x 2 (B/G) x 2 (B/G) x 2 (B/G) = 64 (for each slot there are 2 options B or G) but I would like to understand how to get to this result from the other approaches mentioned above.
But if I were to say, they want exactly 2 Boys and 2 Girls then the slot method fails (or alteast the way I using it).
2 (B/G) x 1 (B) x 2 (B/G) x 1 (G) = 4? It should be 6.
Another type of question is (very similar to the one above):
A committee of 2 has to be selected from a pool made up of male and female members out of 10 candidates (5 male, 5 female). Again same issue how do I calculate the maximum number of possibilities.
Aka, I having trouble grasping the idea where out of a pool there are two types of candidates and we have to make combinations for all of them.
Can you please walk me through how I would use slot, combinations formula and/or anagram for these?
Thanks for your help.
This questions is not specific but a general question about combinatorics, I wasn't sure where to post it, so I hope I posted it in the right forum.
I need some help understanding a Combinatorics concept. I have been using the anagram method and the combinations formula.
I want to understand how to calculate the maximum possible combinations (order does not matter) for questions like:
[Mr. and Mrs. X want to have 4 children...] - At this point how do I calculate all the possible combinations of Boy and Girls they can have if they have 4 children.
One way is the Slot method,
2 (B/G) x 2 (B/G) x 2 (B/G) x 2 (B/G) = 64 (for each slot there are 2 options B or G) but I would like to understand how to get to this result from the other approaches mentioned above.
But if I were to say, they want exactly 2 Boys and 2 Girls then the slot method fails (or alteast the way I using it).
2 (B/G) x 1 (B) x 2 (B/G) x 1 (G) = 4? It should be 6.
Another type of question is (very similar to the one above):
A committee of 2 has to be selected from a pool made up of male and female members out of 10 candidates (5 male, 5 female). Again same issue how do I calculate the maximum number of possibilities.
Aka, I having trouble grasping the idea where out of a pool there are two types of candidates and we have to make combinations for all of them.
Can you please walk me through how I would use slot, combinations formula and/or anagram for these?
Thanks for your help.
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09 Nov 2010, 13:15
Kudos
Bookmarks
First time posting, but I'll take a shot at it. Mods - feel free to correct as needed.
Using the number from above, 64 = total possible outcomes. Now you need to find the number of ways to have exactly 2 boys and 2 girls. You can write them out (BBGG, BGBG, GGBB, GBGB, GBBG, BGGB)
OR
You could use 6C2 to find the number of possibilities. which is (4!)/(2!2!) = 6
That leaves you with 6/64 = 1/16 chance they have exactly 2 boys and 2 girls (in no particular order)
Hope this was helpful. Practice, practice, practice. And just when you think you can't grasp it, practice some more. Don't give up, you'll get it.
Quote:
This can be calculated as 2^4, where 2=possible outcomes (boy or girl) and 4=the number of rounds (babies)
Quote:
I'm assuming you mean to ask, what is the probability of having exactly 2 boys and 2 girls?
Using the number from above, 64 = total possible outcomes. Now you need to find the number of ways to have exactly 2 boys and 2 girls. You can write them out (BBGG, BGBG, GGBB, GBGB, GBBG, BGGB)
OR
You could use 6C2 to find the number of possibilities. which is (4!)/(2!2!) = 6
That leaves you with 6/64 = 1/16 chance they have exactly 2 boys and 2 girls (in no particular order)
Quote:
Similar to above. 10C2 (10!)/(2!8!), which equates 45 possible committee pairings.
Hope this was helpful. Practice, practice, practice. And just when you think you can't grasp it, practice some more. Don't give up, you'll get it.
14 Nov 2010, 20:43
Kudos
Bookmarks
A lot of students get hung up on combinatorics-- they're weird and often less familiar than some of the other GMAT math concepts. Yes, we do need to know the formulas, but pausing before we dig into our formula toolbox can sometimes yield a simpler/less labor-intensive solution. (Although there are certainly times when you *do* need to plunk numbers into a formula and crank them out--FLEXIBILITY is the key to a high score!) That said, there are a couple of different twists operating here.
For the specific question that you posted about 4 children (either boys or girls)--in which order DOESN'T matter--the simplest method would be to skip formulas/slots/anagrams altogether. Because you only have two possibilities for each child, there is a *very* limited set of these possibilities, so it's quite easy to list them out.
-all boys BBBB
-all girls GGGG
-half boys/girls BBGG
-three boys/1 girl BBBG
-three girls/1 boy GGGB
Exactly two boys and two girls in a world where order doesn't matter is even easier-- there is only one situation (two boys and two girls!).
You may have intended to ask this question if order DID matter, which is slightly more complicated. Unfortunately, the slot method may not be as automatically useful here-- the slot method is great when you are choosing from the SAME pool of finite items (like medalists in a race, or anagrams of a single word) but out of 2^4=16 possibilities you have several different "pools" you could be drawing from-- 4 boys, 4 girls, half/half, etc.
****
A really important question to ask of any combinatorics problem is "Am I picking from the SAME pool or DIFFERENT pools?"
****
When you are calculating 16 possible arrangement of boys and girls, you are actually calculating picking from DISTINCT pools 4 times (with the possibility of picking either a boy or a girl being each "pool").
When we ask "how many different ways can I get two boys and two girls?" we are choosing from ONE pool of 4 people--2 boys and 2 girls.
So how many different ways could we arrange the pool of BBGG? 4!/ [(2!)(2!)] = 6-- this is like a "how can you rearrange the letters in the word PIZZA"-style anagram problem. To do this, we calculate the permutations (order matters!) by taking the factorial of the number of elements (4), then divide by the factorial (or in this case, factorials) of any repeated element(s) (2 boys and 2 girls). The slot method would be the same--we just have to remember to still divide by the factorials of the repeated elements, and we must remember that our "pool" consists of 4 elements (two boys and two girls).
A just-as-fast method for this particular questions, however, may be listing out the possibilities. You are dealing with only 16 possible outcomes (you may have made a careless error in your calculation above...2^4 is only 16), and if you are looking for an even smaller subset: 2 boys and 2 girls, listing them out is fairly straightforward.
GGBB
BBGG
BGGB
GBBG
BGBG
GBGB
...just remember to be organized when you do so. Notice here that each possibility has an opposite "pair"
If we are asked for the PROBABILITY of getting two boys and two girls, in which order does not matter, then we need to calculate both the total number of possible outcomes (2^4=16) and the number of ways in which we could get 2 boys and 2 girls ( 4!/ [(2!)(2!)] = 6). Our answer is 6/16, or 3/8.
* Notice that although we said order does *not* matter for this probability problem, we use the number we got for the permutation sub-problem, in which order *does* matter. 2^4=16 gives us the total number of gender-arrangements in which to have 4 children, where order does matters, so we must use the number we calculated for the ways to get 2B/2G in which order does matter (6) by 16 to get the probability: the # of outcomes we want / # total outcomes.
******
The anagram METHOD, or anagram GRID, is most useful as a way to think through combinatorics problems when you are choosing SOME, but not ALL members of a group. For example, if you were going to assign gold, silver, and bronze medals to 3 out of 10 runners, the anagram would look like this:
RUNNER: | A | B | C | D | E | F | G | H | i |
MEDAL: | 1 | 2 | 3 | S | S | S | S | S | S |
...where the S's correspond to the slower runners we don't care about. It's like figuring out the ways to arrange GGBB above, but our repeated elements here are the 7 S's-- order matters, so our formula is the same: the number of elements is 10 and the number of repeated elements is 7: 10! / 7! = (10*9*8)=720
The SLOT method is exactly the same thing: 3 slots, 10 possibilities for the first, 9 for the second, and 8 for the third: 10*9*8=720 All you're doing with the slot method is executing the canceling at the front end rather than the back.
You can still use the Anagram grid if you are choosing all elements, but it's unnecessary-- if that's the case you have no repeated elements, so you aren't dividing by anything, and you have a straight factorial. To calculate the number of ways to arrange 10 runners, you would calculate 10!.
To be perfectly honest, I don't emphasize the anagram method in my classes-- the upside to this method is that you can apply it across many different kinds of problems, but the downside is that it confuses some students and can be very labor intensive, while the slot method is intuitive and quicker. BUT you must know the conditions in which you can use ANY method (for example, we had to do a little thinking about the boy/girl problem you gave to figure out what the *real* problem was...only then could we use our slots/anagrams/formulas). Ideally you will know how to use one visual method (slots or anagrams) AND the formulas.
I repeat: the slot method is *the same thing* as the formula for combinations and permutations--it's simply a shorthand way of doing the canceling across the division bar up front rather than later in the process. Be wary of applying any method automatically. Usually, when there is a twist, a short pause to reason through the path to the solution is a really smart move before applying *any* method-- slots, anagram, formula, or simply listing out possibilitities.
This post is getting long so I'm going to pause here and will come back to answer your second q in a separate post.
For the specific question that you posted about 4 children (either boys or girls)--in which order DOESN'T matter--the simplest method would be to skip formulas/slots/anagrams altogether. Because you only have two possibilities for each child, there is a *very* limited set of these possibilities, so it's quite easy to list them out.
-all boys BBBB
-all girls GGGG
-half boys/girls BBGG
-three boys/1 girl BBBG
-three girls/1 boy GGGB
Exactly two boys and two girls in a world where order doesn't matter is even easier-- there is only one situation (two boys and two girls!).
You may have intended to ask this question if order DID matter, which is slightly more complicated. Unfortunately, the slot method may not be as automatically useful here-- the slot method is great when you are choosing from the SAME pool of finite items (like medalists in a race, or anagrams of a single word) but out of 2^4=16 possibilities you have several different "pools" you could be drawing from-- 4 boys, 4 girls, half/half, etc.
****
A really important question to ask of any combinatorics problem is "Am I picking from the SAME pool or DIFFERENT pools?"
****
When you are calculating 16 possible arrangement of boys and girls, you are actually calculating picking from DISTINCT pools 4 times (with the possibility of picking either a boy or a girl being each "pool").
When we ask "how many different ways can I get two boys and two girls?" we are choosing from ONE pool of 4 people--2 boys and 2 girls.
So how many different ways could we arrange the pool of BBGG? 4!/ [(2!)(2!)] = 6-- this is like a "how can you rearrange the letters in the word PIZZA"-style anagram problem. To do this, we calculate the permutations (order matters!) by taking the factorial of the number of elements (4), then divide by the factorial (or in this case, factorials) of any repeated element(s) (2 boys and 2 girls). The slot method would be the same--we just have to remember to still divide by the factorials of the repeated elements, and we must remember that our "pool" consists of 4 elements (two boys and two girls).
A just-as-fast method for this particular questions, however, may be listing out the possibilities. You are dealing with only 16 possible outcomes (you may have made a careless error in your calculation above...2^4 is only 16), and if you are looking for an even smaller subset: 2 boys and 2 girls, listing them out is fairly straightforward.
GGBB
BBGG
BGGB
GBBG
BGBG
GBGB
...just remember to be organized when you do so. Notice here that each possibility has an opposite "pair"
If we are asked for the PROBABILITY of getting two boys and two girls, in which order does not matter, then we need to calculate both the total number of possible outcomes (2^4=16) and the number of ways in which we could get 2 boys and 2 girls ( 4!/ [(2!)(2!)] = 6). Our answer is 6/16, or 3/8.
* Notice that although we said order does *not* matter for this probability problem, we use the number we got for the permutation sub-problem, in which order *does* matter. 2^4=16 gives us the total number of gender-arrangements in which to have 4 children, where order does matters, so we must use the number we calculated for the ways to get 2B/2G in which order does matter (6) by 16 to get the probability: the # of outcomes we want / # total outcomes.
******
The anagram METHOD, or anagram GRID, is most useful as a way to think through combinatorics problems when you are choosing SOME, but not ALL members of a group. For example, if you were going to assign gold, silver, and bronze medals to 3 out of 10 runners, the anagram would look like this:
RUNNER: | A | B | C | D | E | F | G | H | i |
MEDAL: | 1 | 2 | 3 | S | S | S | S | S | S |
...where the S's correspond to the slower runners we don't care about. It's like figuring out the ways to arrange GGBB above, but our repeated elements here are the 7 S's-- order matters, so our formula is the same: the number of elements is 10 and the number of repeated elements is 7: 10! / 7! = (10*9*8)=720
The SLOT method is exactly the same thing: 3 slots, 10 possibilities for the first, 9 for the second, and 8 for the third: 10*9*8=720 All you're doing with the slot method is executing the canceling at the front end rather than the back.
You can still use the Anagram grid if you are choosing all elements, but it's unnecessary-- if that's the case you have no repeated elements, so you aren't dividing by anything, and you have a straight factorial. To calculate the number of ways to arrange 10 runners, you would calculate 10!.
To be perfectly honest, I don't emphasize the anagram method in my classes-- the upside to this method is that you can apply it across many different kinds of problems, but the downside is that it confuses some students and can be very labor intensive, while the slot method is intuitive and quicker. BUT you must know the conditions in which you can use ANY method (for example, we had to do a little thinking about the boy/girl problem you gave to figure out what the *real* problem was...only then could we use our slots/anagrams/formulas). Ideally you will know how to use one visual method (slots or anagrams) AND the formulas.
I repeat: the slot method is *the same thing* as the formula for combinations and permutations--it's simply a shorthand way of doing the canceling across the division bar up front rather than later in the process. Be wary of applying any method automatically. Usually, when there is a twist, a short pause to reason through the path to the solution is a really smart move before applying *any* method-- slots, anagram, formula, or simply listing out possibilitities.
This post is getting long so I'm going to pause here and will come back to answer your second q in a separate post.
