Hi Monica, glad the article helped! In this case, it depends on whether the order of doing the circuit matters: is “pushups, jumping jacks, squats, lunges” the same as “lunges, squats, jumping jacks, pushups”?
If the order doesn’t matter, you can just multiply the choices available in each category:
4 choices in first category * 4 choices in second category * 4 choices in third category * 4 choices in fourth category = 256 choices
If the order does matter, we need to see how many ways we could re-arrange the categories. This is the number of permutations of 4 categories, or:
4 choices for category coming first * 3 choices for category coming second * 2 choices for category coming third * 1 choice for the last category = 24 category orders
So the total workouts, when category order matters, is 24 * 256 = 6144
Thank you very much for the intersting topics I am a big fan of your website.
I just have a question:
In European Champions League, There is a draw takes place in March every year to define what teams would play against each other among the qualified teams.
This year, we have 8 qualified teams:
Barcelona
Bayern Munich
Real Madrid
Monaco
Porto
Paris Saint Germain
Juventus
Atliteco Madrid
The draw will define the next 4 matches in this championship, The order is important here,
so “Barcelona versus Bayern Munich” (This match will take place in Barcelona) differs from “Bayern Munich versus Barcelona” (This match will take place in Munich)
The result of the draw would be like:
Barcelona versus Porto
Bayern Munich versus Monaco
Real Madrid versus Juventus
Atliteco Madrid versus Paris Saint Germain
The question is, how many possiblities can we have as a result for this draw?
My answer is (correct me if I am wrong):
P(8,2)*p(6,2)*p(4,2)*p(2,2) = 40320 (but this is really a huge number!!)
I’m trying to help my son with math and i am stuck… with a total of $2.00 in coins how many different combinations are possible using a .50-cent piece, .25-cent piece and .10-cent piece ??? please help
Hi. I could not understand one point. You said choosing 3 people from a group of 10 is a combination but choosing President, VP and waterboy is permutation. Why is that? How does it matter who is selected first? How do we know which person of the three selected person gets which post? Please explain as this topic bothers me a lot. Thanks.
So if I understand this correctly…
r-permutations are really the total permutations divided by the permutations of what it is not.
ex: P(10,3) ways to choose 1-3rd place out of 10 people
total perm = 10! = 3628800
perm of those we are not counting ( 7!) = 5040
total/perm of not counting = 3628800/5040 = 720.
and combinations are that number (720) divided by permutations of different ways to order 3 (3! = 6)…
thus 720/6 = 120 = C(10, 3).
I know this is convoluted but I’m trying to grasp this at fundamental level because I dont understand bit strings. How many bit strings of length 10 have four 1s. supposedly the answer is C(10,4) but I cant find an explanation of why that is. the 10 in that scenario I thought was reserved for the size of the set or the number of ways on thing could be chosen like C(52,5) for cards. _ _ _ _ _. the number of choices for 1st space is 52, then 51 etc. and 5 represents the number of combinations we are counting.
how does the fact that I only have 2 choices for each digit come into play ( in the bit string example). total number of permutations I thought would be 2^10 = 1024 and number of r-permutations (r = 4) would be …I think im actually losing myself here. Maybe I missed it as I was scrolling through comments, but could you explain combinations or permutations of bit strings. and perhaps how the choice of only 1 or 0 comes into play there? you can use the example of bit string length 10 choose 4 if you want or why i use C(10, 4). supposedly being (10!/6!)/4!= 210. Why is it 1098… int the numerator. there arent 10 choices * 9 choices… im lost and I love your work by the way. I have already recommended your site to fellow students.
To win at LOTTO in one state, one must correctly select 7 numbers from a collection of 47 numbers (1 through 47). The order in which the selection is made does not matter. How many different selections are possible?
I know to use the formula for a combination and I can determine the values of n and r…but i get stuck at simplifying to get the correct answer…
I have 6 items - 1A, 1B, 2A, 2B, 3A, & 3B. I’m trying to develop a formula to determine the number of possibilities if 1A and 1B, 2A and 2B, and 3A and 3B cannot exist together. I need groups of 1 (the answer there is 6); groups of two (where 1A-2A is legal but 1A-1B is not-- Should total 8) and finally three (1A-2A-3B is legal but 1A-1B-2A is not- Should total 8). I worked it out graphically but if the number of items is 20 instead of 6 it turns into a graphic nightmare. And to take is one step further lets add 4 and 5 (without A or B) and up to 5 possible items.
Hi Kalid, this really helped me a lot! Thank you for that. But about the question how many different numbers above 6000 can be formed from the digits 3,4,5,6,7…
The question seems a little ambiguous to me. It doesn’t say the number has to have 4 digits, just that it has to be greater than 6000. Also, it doesn’t say whether or not the digits may be repeated. The way I understand this is that it is therefore any number less than infinity and greater than 6000, so long as it has those 5 digits. Please help!
Thanks.Now I got what really combination means.Could you explain why if you take 3 heads out of 5 coin tosses your value of k!=3!=6 while p(5,3)=60.I’m confused why there is only 6 variants since for each one of the 60 outcomes you could change it for 3 heads and 2 tails so it would be 60*3!*2!=720?I took heads as H1,H2,H3 and tails as T1,T2.To place heads in 3 positions there are 6 ways and to place tails in 2 positions there are 2 ways.So for 60 outcomes it would be 720 ways.Could you help me.
Considering a coin flipped three times and wanting 2 heads, in what sense doesn’t order matter that gives us 3 combinations? What are the six permutations?
can someone help me please…
Malini decides to select 5 songs from a list of 10 songs to be sung during a concert. in how many ways can she
a) make a selection
b) presents the songs
Thanks for your great explanations. I understood the combinations but couldn’t find good explanations of the notation. For the 2nd question above (the one about pulling 2 face cards from a deck,) could you work it out completely? Also, for the 3rd question above (the one about forming numbers larger than 6000,) wouldn’t you also have to add the five digit-numbers which you could make? Like 34, 567, etc.? Thank you!
I am a big fan of your website.
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