How many different sets of exactly four people can be selected from a group of six people?
815499
The correct answer is 15, but the proper formula is (6C4)/2\n\nBy the way, you can check this by figuring out how many groups of 2 you can have, since every group of four being chosen is the same as a group of two not being chosen.  For the first person, there are six people to choose from.  For the second, there are five.  However, since picking Person I and Person II is the same as picking Person II and Person I, you have only half as many possible combinations, or 15.
1137758
1) Let's first consider not the combination but the permutations without repetition (considering the order).\n\nSuppose to have 39 numbered cards in your hands and shuffle them well. \nIf you get one card and you drop it on the table you have 39 different possibilities, obviously.\n\nNow, from the remaining 38 cards in your hands, If you drop another card on the right side of the previous card, you'll have 38 different possibilities for just that second choose, but If you consider the order of both cards you'll have exactly 39*38 = 1482 different possibilities that the 2 card will be ordered and choosen in that way.\n\nIf you get a 3rd card from the remaining 37 in your hand you'll have on the table in the right side of previous exraction, a group of 3 ordered cards choosen by 39*38*37 = 54834 different ways will be on the table.\n\nRecursively go to choose until 6 ordered cards will be the table and you'll have 39*38*37*36*35*36 = 2349088560 possibile ways if you consider the order. \n\n2) Now let's see what happend when the order of the selection matters as you asked in your question. \nSo let's make this consideration:\n\na) If you have one card (A) you don't mind about ordering problems:  {A} = (A)\n\nb) if you have two cards (A, B) you have to consider the couple (A, B) = (B, A) so we write {A, B} are commutative;\n\nc) if you have three cards "A", "B" and "C" you can group (A, B) as one element {A, B} because order doesn't matter (point b).\n\nSo we can consider the 3 cards (A, B, C) as:\n ( {A, B}, C ) = ( C, {A, B} ) as we made in the point b.\nYou could also group (A, {B, C}) = ( {B, C}, A)  or {A, C} and then swapping with B.\n\nSo with 3 elements we have 3 different ways to create couple to swap so we can create 3*2 = 6 different combination.\n\nIf we have 4 cards we know it's possible to choose 4 different ways to create list of 3 cards. Let's stop one each time with your finger and you can see that you can group easily the remaining other 3's.\nSo the possible combination with 4 cards is 4 times the possible combination I have with 3 cards. So it's  4 * (3 * 2).\n\nWith 5 cards you'll have 5 times the number of combinations we found with 4 cards, so it's  5 * (4 * 3 * 2).\n\nFinally, with 6 cards in a table arrange in 6 * 5 * 4 * 3 * 2 = 720 different ordered ways.\n\nNow, your question was how to do if the order of selection matters and in the previous example we find that the number of possibilities we have when we select 6 cards ordered is this number 39*38*37*36*35*34 = 2349088560. And we know now that this number is 720 times greater than the possibilities wether the order of selection doesn't matter. (720 = 6 * 5 * 4 * 3 * 2).\n\nIn other words, 39 * 38 * 37 * 36 * 35 * 34 is the number of permutations without repetition of 39 elements grouped by a list of 6 elements considering the order.\n\nAnd now we know with 6 elements you can have 6*5*4*3*2 possible different positions of those elements. \n\nSo Just divide 39*38*37*36*35*34  by 6*5*4*3*2 and you'll have the result of the combinations.\n\nSo to see how many combination of 6 cards bye a set of 39 just divide 39*38*37*36*35*34 / 6*5*4*3*2 = 3262623.\n\nthe number 6*5*4*3*2 could be wrote "6!" using the the factorial function.\n\nThe factorial function of positive integer n, is defined as the product of first n numbers:\n\nn! = 1 * 2 * 3 * ... * n\n\nand is a convention that the product of no numbers at all is 1.\n\n0! = 1\n\n\nThe factorial is also defined recursively by the follow definition:\n\n(n+1)! = (n+1) * n!\n\n\nSo if you have n object and you want to know how many permutation (the order matters) without repetition (each object can be chosen only once) the objects makes with a number k of selection the formula is:\n \n  n! / (n-k)!\n\nso it's the product of k terms starting from n and descending until reached (n-k+1).\n\nn*(n-1)*(n-2)*.
795663
Jesi has the right answer...\n\nIf there were ten, how would they reach decisions without the possibility of a tie vote? There are nine.
198261
The least number would be 6 but in case someone is sick or can't play you might want to have substitutes. Also if you have enough for two teams that's nice because in practice you can see how you play against another team. This can help you play better. But really all you need is 6 people.
420921
No more than four please consider that no more than two each per humanoid. Allowances are made for the handicapped and the well endowed.
889116
This is a combination problem, because the order of people doesn't matter. In permutations, the order does matter.\n\nThe formula for the number of combinations of k items from a set of n items total is:\n\nn! / [ k! (n - k)! ] where ! means factorial (n * (n-1) * (n-2) ... * 1)\n\nIf your case that would be 35! / (8! * 27!). Sounds like a calculator problem to me.\n\nI get 23535820 combinations.\n\nGood luck on your homework!
617746
sample space {1,2,3,4,5,6}
840938
The number of handshakes is 302C2, i.e. (302)(301)/2 = 45451.\n\nYou're essentially looking for the total number of pairs of people in the group.  With your answer of 90902, I suspect that you were on the right track.\n\nSo to find the number of pairs, you have 302 choices for the first and then 301 choices for the second, making 90902 possibilities.  We have to cut that in half because this is counting 'pick person A then pick person B' and 'pick person B then pick person A' as two distinct choices, but this is really just two ways of looking at the same handshake.  Cheers!
367272
Suppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted. Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = P(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!]. \nNow you should be able to derive it whenever you get confused.
233027
There are 8! different arrangements (permutations) of the 8 friends.\n\nThere are 4! different arrangements of the 4 boys.\n\nThere are 5 different locations where the leftmost boy can sit when the 4 boys sit together.\n\nTherefore, there are 5*4! (or 5!) arrangements of the boys where all the boys sit together.  But there are also 4! arrangements of the 4 girls in the remaining slots.  So there are 5!*4! arrangements where the 4 boys sit together.\n\nSince there are 8! arrangements overall, the probability of all boys sitting together is (5!*4!)/8!.\n\nSimilarly, if the boys sit to the left and the girls sit to the right, then the probability is (4!*4!)/8!.\n\nInterestingly, this last answer is also the same probability as boy-girl-boy-girl-boy-girl-boy-girl and also the probability of girl-boy-girl-boy-girl-boy-girl-boy.  Therefore, the probability of alternating seats is (2*4!*4!)/8!.\n\nRecall that n! (read n-factorial) is n*(n-1)*(n-2)*...*3*2*1.
230780
1,2,3,4,5,6  Try those!  (Same odds as any other combination of numbers!)
917858
Only 1, 3, and 4 are allowable for the hundreds digit.  Anything goes for the rest.  So there are 3 choices for the hundreds digit, 5 choices for the tens digit since you can't use whatever you cose for the hundreds, and 4 choices left for the ones digit.\n\n3*5*4 = 60
631878
If you are only concerned about the ways that you can place the 15 women between the 10 men, then you have 9 positions to place the 15 women.  (You don't have to worry about the arrangements for the men since there has to be one in between there is only one way to arrange.) These are the ways to do it (regardless of order).\n\na) 1 1 1 1 1 1 1 1 7\nb) 1 1 1 1 1 1 1 2 6\nc) 1 1 1 1 1 1 1 3 5\nd) 1 1 1 1 1 1 1 4 4\ne) 1 1 1 1 1 1 2 2 5\nf) 1 1 1 1 1 1 2 3 4\ng) 1 1 1 1 1 1 3 3 3\nh) 1 1 1 1 1 2 2 2 4\ni) 1 1 1 1 1 2 2 3 3\nj) 1 1 1 1 2 2 2 2 3\nk) 1 1 1 2 2 2 2 2 2\n\nNow to rearrange these, you would have to use the multinomial coefficient.  You have to use this because you have repeats: For example in a), you can switch two 1's around and get the same result, but moving the 7 will give you a different result.\n\nFor a), there are 9!/8!1! = 9 ways to arrange.\nFor b) and c), there are 9!/7!1!1! = 72 ways to arrange.\nFor d), 9!/7!2! = 36 ways.\nFor e), 9!/6!2!1! = 252 ways.\nFor f), 9!/6!1!1!1! = 504 ways.\nFor g), 9!/6!3! = 84 ways.\nFor h), 9!/5!3!1! = 504 ways.\nFor i), 9!/5!2!2! = 756 ways.\nFor j), 9!/4!4!1! = 630 ways.\nFor k), 9!/3!6! = 84 ways.\n\nSo in total, there are 9 + 72 + 72 + 36 + 252 + 504 + 504 + 84 + 756 + 630 + 84 = 3003 ways to place the 15 women between the 10 men.\n\nNow if after that you want to know how many ways there are to arrange if you do make a difference between men and between women, then multiply by the number of ways to arrange the men and women.\n\n3003*10!*15! = 14,250,102,106,034,995,200,000 ways.\n\nNotice this is less than the number of ways to arrange all 25 people regardless of gender.  That would have been 25! = 15,511,210,043,330,985,984,000,000.
149355
omg your kidding me... thats a trick question. the set would already  be over if it is 6-4.
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For the first question, that means you want to know about committees with exactly 3 women and committees with exactly 4 women. Since these are independent cases, we consider each separately and add the result. \nFor 3 women, we need to know how many unordered groups of 3 women we can choose from 4 women, so we have 4[Choose]3. The men have 2 people on this committee, so they contribute 4[Choose]2 to the product for a total of (4[Choose]3)*(4[Choose]2) committees with exactly 3 women on board. \nThere is only one way to grab an unordered set of 4 objects from a set of 4 objects, so we go right to the men. There are 4 ways the men can fill the last spot, so this is 1*4=4 different committees with 4 women on board.\nThus the total number of committees with a majority of women on board is (4[Choose]3)*(4[Choose]2) + 4.\nThe second part is very similar to the first.
913636
if there were 12 groups each with 1 cheerleader in....you will have 12 groups
276799
happy family=4-5 members. possibly 6