Great post.
Pls i nid a help
In how many ways may twelve persons be divided into three groups of 2,4 and 6 persons
the answer is 13860 but how did we get dat…
thanks.
This post is great, but IMO in dealing with permutations you jumped to a “mechanical” explanation instead going further with the intuition path. Saying “We only want 8 * 7 * 6. How can we “stop” the factorial at 5?” and "And why did we use the number 5? Because it was left over after we picked 3 medals from 8. " is not very intuitive to me. The way I would interpret it: “Hey, I initially have 8!=8x7x6x5! ways of giving 8 medals to 8 participants, where the order matters. This means that from every one of 8x7x6 way to give the first 3 medals to 3 participants, there are 5! ways of giving the remaining 5 medals to the remaining 5 participants (multiplication rule). But we are not really interested in how we distribute the medals among those remaining 5, so we focus on the first three medals, ending up with the initial 8x7x6 ways of distributing the first 3 medals. Of course, the mechanic translation of this is the formula you correctly state”.
Keep up the good work, Kalid!
You really make it seem so easy along with your presentation however I in finding this topic to be really one thing that I feel I might by no means understand. It sort of feels too complex and extremely wide for me. I am having a look ahead for your next post, I will attempt to get the hold of it!
C (44,6) how do i show work with these big numbers? i need to show my professor that i know how to work it out without typing it in to a calc…
Hi Ruben, that’s a great perspective/analogy, thanks for sharing!
I have enjoyed reading your answers. You make a lot of sense and I appreciate it. However now I have a question if you can help me. In how many different ways can you select a committee of 3 people from a group of 13 memebers: The committee members consist of a chairperson, treasurer, and a secretary? I think it is a combination because it is not asking you to specifically pick who will be the c ,t, or s. It is just asking to select a committee. Is this correct? Please help.
My teacher explained it to me, but i am still confused! If there are 8 sodas and you want to make 3 combinations out of it. How many combinations could there be?
Hi Tracey, great question. Actually, this is still a permutation because re-arranging the roles would matter (i.e., keeping the same people but switching roles would count as a different “group”). A better intuition might be “permutations are different if you re-arrange your choices, combinations are the same”.
So picking 3 people to be in a group is a combination, but picking roles in the group is a permutation. It’s really tricky, and I still mix it up. If it makes it easier, combinations work when every choice is “identical” (i.e., all the roles are the same), and permutations work when there’s something different between the roles.
So, that said… this is very similar to the “gold, silver, bronze” question above, but picking 3 from 13 (instead of 3 from 10).
Thanks a lot for the simple approach Kalid! I can understand more how it is like this now 
Thank you!
I think it would be even better if later on you could also explain how the other permutation rules work, like circular permutations, what if two people ought to sit together how do you arrange them, and the like :)) Wonderful work Kalid! 
This is my first time visit at here and i am actually pleassant to read all at one place.
this is also my first time visit at this website absoutely it help me …
4 golfers toss a tee to decide the order they will tee off (a tee is thrown in the air). whoever whoever the tee points to gets to go first. then the 3 remaining do it, and so on… what is the probability that they tee off in alphabetical order?
the answer is 1/24… if you could tell me why, it would be golden.
It’s very easy to find out any matter on net as compared to
textbooks, as I found this post at this website.
From a group of 7 men and 6 women, 5 people are to be selected to be assigned to a committee in such a manner that at least 3 men are assigned to the committee… How men ways are there to do so?? … Please help… Thank you!!
Thanks Ari!
There are 3 sets of twins and 6 tickets. Two tickets are for row A, 2 Row B and 2 for Row C. Each twin selects a ticket randomly and sits in it that seat. What is the probability that each twin sits in the same row as his/her own twin? Dont even no where to begin…help!
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