10 pairs of shoes are well mixed up.4 shoes are randomly picked. What is the probablity of getting at least 1 complete pair
This is really an excellent way of explaining the things!!!
Hi Pras, thanks for the comment!
@Sahil: It’s a good question, but I want to make sure I’m not doing someone’s homework for them :).
In general, it’s easier to find the chance of “zero matches” and subtract this from 1, vs. finding the chance for 1 match.
So let’s find the chance for zero matches. Imagine picking your first shoe, A: nothing special here, you aren’t going get the match on a single shoe.
You pick shoe B: You have a 18/19 chance of getting zero matches (only A’s partner would match, of the 19).
You pick shoe C: You have a 16/18 chance of zero matches (only A and B’s partner would match, of the remaining 18).
You pick shoe D: You have a 14/17 chance of not getting any matches (only A and B and C’s partner would match, of the remaining 17).
If you multiply these chances you get the total chance for zero matches. Subtract from zero to get the chance for any match. At least I think that’s how it goes 
hey
I was about to crazy solving this sum which i now find was actually so simple thanks
I feel the answer to this is simple but i am just not able to get it…
In an examination there are three multiple choice questions and each question has 4 choices. The number of sequences in which a student can fail to get all answers correct is…
Hi Safa, for that question it helps to take it one step at a time.
In a test with only 1 question, how many ways can you be wrong? 3. (Suppose the right answer is D… you could answer A, B or C).
Now how about 2 questions? Well, you have 3 ways to get the first question wrong, and another 3 ways to get the second one wrong. So the total is 3 * 3 = 9. (Let’s say the right answer is D and D. Then AA AB AB BA BB BC CA CB CC are all wrong).
Similarly, if you have 3 questions, then there is 3 * 3 * 3 = 3^3 = 27 ways to get all answers wrong. (You can write it out but will take a while: AAA AAB AAC ABA ABB ABC ACA ACB ACC… you get the idea )
Hope this makes sense,
-Kalid
thanks too much
i understood it
thanks again
Nice explanation. I was wandering if you could explan some more about COUNTING…
Thanks…
splitline…
Hi Splitline, thanks for the suggestion – I may cover counting in the subject of an upcoming article.
At a high level, to count the number of ways to do something, you multiply all the choices together. So, if you want to count how many ways to get 3 cards in poker, you’d do 52 (first option is to pick any card) times 51 (second option is any of the remaining cards) times 50 (third option is any of the leftover cards).
This is the general idea – a full article may be needed to make it more clear.
It was a very useful site,indeed!
Thanks Geetha!
Hi
I would appreciate if you answer this:
A survey question has 6 answers, you can choose a single answer or any combination from the 6. How many possible combinations are there?
Thanks a lot
Hany
Hi Hany, this question is a bit different. In this case you have 2 choices (use or don’t use the question), and you make this decision 6 times. So the number of possibilities is
2 * 2 * 2 * 2 * 2 * 2 = 2^6 = 64.
Update: A great point was made below (comment #163) – the case of ZERO questions should be removed. So you have 64-1 = 63.
Thank you very much Kalid, you made my day!
What is the exact term for this type of calculation?
Now, I want to put a formula for this in Excel to automatic coding of the 64 possibilities; is there a way to do that?
Thanks a lot
Hany
didn’t help, im more confused
Sorry it didn’t work for you – try to forget it if you’ve become more confused :).
This is more of a refresher for people that learned combinations and permutations but then later forgot the formula [like me]. If you’re learning this in class, try running through a few examples in your textbook.
Please help! How many combinations of wins are there in 12 football games?
Is this part of a field called ‘combinatorics’ or is that something totally different?
If so, could you do another explanation in that field, I have been reading your posts and this one and the ones on e and ln are terribly interesting.
@Kathleen: I’m not sure if I understand the question.
If counting the number of sequences [Win-win-win-lose-lose…], you can win or lose each game. You have 2 choices at each game, and 12 games, so there are 2^12 = 2048 possibilities total.
If you’re counting the number of different records (6-6, 12-0, etc.) then there are only 13: 12-0, 11-1, 10-2 … 1-11, 0-12 [it’s 13 because we’re counting down to 0, not 1].
@Jon: Glad you liked the articles! Combinatorics is about the number of ways to “count” something (from the wikipedia article), so permutations and combinations would fall under that title.
Permutations/Combinations also occur in statistics, when you try to find the likelihood of a certain event happening out of all possible events [and you need to count the number of possible events].
Given the counting questions here, I’ll add another combinatorics article to my topic list
Thank you very much Kalid.
I started as a philosophy major, and decided to go into computer science/A.I. for my master’s where I have been discovering an unexpected love for the beauty of mathematics. And it makes me smile to see sites like this one with open forums and quick feedback for interesting topics. Thank you again, and this will definately be a site I check regularly!