Permutations/ combinations

[dsony2284]

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I am having problems realizing when to use the combination / permutation equations. Does permutation involve some sort of order? while combination questions merely ask for different combinations that are possible.Can someone please explain what to look for maybe in using which formula?

Also....

For those of you with achiever, on the QR test 1 there is a question that asks What will be the likelihood of having exactly 1 boy in a family planning for three children?

i dont understand the permutation used here and why..can someone explain this thanks

Thanks...

 

[jay47]

Ok, an easy way that I learned to do probability is to use this general formula-

(number of ways to do "x")/(total possible number of outcomes)

for example-

How many ways can you have one boy in three tries?("x" in this case )
You can have one on the first try OR one on the second try, OR one on the third try. This is entered as a permutation, 3 nPr 1=3. In a permutation order does matter. For example, if it asked, how many ways can two boys be born out of three children, it would matter if boy A was born before boy B, considering those are their names. If boy A is born second, that is a different event than before. Thus you must use a permutation for the first case and in my example. However, for the 1st example, it does not matter which you use because 3 nCr 1= 3 nPr 1. This is just coincidence though. For more go to-

http://mathcentral.uregina.ca/QQ/database/QQ.09.00/ashleigh1.html

And...

How many different total possibilities are there for 3 children? 2^3=9


Combinations could be used for something like "How many ways can you draw 5 cards out of a 52 card deck?" In each hand of 5, the order of cards drawn does not matter because they are considered a group, not individuals.

Someone please correct me if I am wrong...I need to know too...lol
So I believe that answer to this problem should be 3/9=.33

 

[Sillygirl]

hehe does anyone else think its funny that we're trying to figure out the same problem that a 10th grader in algebra 2 is trying to figure out?? haha

 

[z3u2]

Someone please correct me if I am wrong...I need to know too...lol
So I believe that answer to this problem should be 3/9=.33

is this the question u were answering?
What will be the likelihood of having exactly 1 boy in a family planning for three children?

if so, then ur answer is incorrect.
three ways to do it:
1. (3 choose 1) * (0.5)^1 * (0.5)^2 = 0.375

2. total number of boy/girl combinations = 2 * 2 * 2 = 8
Total number of outcome that fits the answer = 3 (have a boy as the first baby, second babay and third baby)
probability = 3/8 = 0.375

3. list out all the possibilities:
first - second - third
1. girl - girl - girl
2. girl - girl - boy
3. girl - boy - girl
4. girl - boy - boy
5. boy - girl - girl
6. boy - girl - boy
7. boy - boy - girl
8. boy - boy - boy
result 2, 3 and 5 are the desired ones.
probability = 3 / 8 = 0.375