The toys are all identical. So, all you are trying to do is to choose the NUMBER of toys that get put in each box. The condition given says that you must put at least one toy in each box; so, you have 8 identical toys remaining, and can distribute them between the three boxes in any way you like. This can be thought of using a 'stars and bars' argument: any arrangement of these 8 toys is equivalent to a sequence of 8 $*$'s and 2 $|$'s, where each $*$ represents a toy and the $|$'s represent the cutoffs between boxes. For instance, $$ **|*****|* $$ represents the first box getting two of these eight toys, the second getting five, and the third getting one. The number of such sequences is $\binom{10}{2}$: there are 10 positions total, and we need to choose which 2 of those positions are bars. Statement I is correct, Statement II is correct; Statement II is not a correct explanation for Statement I Statement I is correct, Statement II is incorrect Statement I is incorrect, Statement II is correct Statement I is correct, Statement II is correct; Statement II is a correct explanation for Statement I Open in App Suggest Corrections 0 First of all, this is a question that uses the "stars and bars" technique. In case you don't know "stars and bars", we can think of the problem as laying out the 10 balls in a row and then building boxes around the balls. Since the two walls "at the end" of the boxes is trivial, we ignore them and look only at the walls that actually divide the balls. We can denote the balls with a 0 and the walls of the boxes as a 1. We can have this: 0001000100010 which is 3 balls in three boxes and 1 ball in one box. And so what we can do is look at the number of ways we can distribute the walls (the 1s). This is a combinations problem, the formula for which is: #C_(n,k)=(n!)/((k!)(n-k)!)# with #n="population", k="picks"# And so with "stars and bars" problems, we look at the number of unique combinations of placing the 0s and 1s. 1. no empty boxes In this first case, we can assign 4 balls and put one each into a box. That leaves 6 balls to be divided amongst the 4 boxes. Since we need 3 box walls to denote the 4 boxes (just like in our example above), we can then find unique combinations of 6 balls and 3 walls across 4 walls (using 3 walls) is: #((6+3),(3))=((9),(3))=(9!)/((3!)(6!))=(9xx8xx7)/6=84# 2. at most 3 empty boxes With the 3 wall technique, we'll always have at least one box with balls (we'd need a 4th wall to wall off the boxes completely from the balls. One arrangement is: 1110000000000 this is the first 3 boxes with no balls and the last with all of them) And so we can look at the number of ways to combine all 10 balls and 3 walls: #((10+3),(3))=((13),(3))=(13!)/((3!)(10!))=(13xx12xx11)/6=286# ~~~~~ This question/answer on stackexchange.com was invaluable in helping me arrive at a methodology: https://math.stackexchange.com/questions/1083000/stars-bars-question-identical-balls-in-distinct-boxes |
In how many ways 10 identical toys are placed in 4 distinct boxes such that no box is empty
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