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There are 3 scholarships of unequal value.In how many ways can they be distributed among 10 students giving not more than one to a student?

how do you go about this question?it seems to be confusing......

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Jim S. | Physics (and math) are fun, reallyPhysics (and math) are fun, really
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       You have 10 people to choose from for the first scholarship, Now one person has one scholarship and there are 9 people left and 2 scholarships so there are 9 ways to give out the second and 8 ways to give out the third. So the total is 3*10*9*8=2160 ways to distribute the three different scholarships.


         After thinking about this more the confusing part is you are not asked how many ways can three students be chosen out of 10 that would be 10C3=120 but how many ways can 3 different scholarships be given to 10 students.   I would say the answer is 10*3 or 30.
But why should the answer be different with the first one?
Does it change in relation to the method you use or what?
       The short answer is because the first one is incorrect. The factor 10*9*8 is actually the number of permutations of 10 items taken 3 at a time. A permutation is counted as long as there are three elements. So that means, if we numbered the students 1,2,….10, that in some of the groups of three there would be {1,1,1} and {2,2,7} and so on, these groups can not be real because there is only one student number 1. So by going at the problem in this manner I was counting groups that could not exist. That was dumb on my part and I apologize if I confused you.
    The way to think about this problem is as follows: we can choose a student in exactly 10 ways and for each student chosen we can select a scholarship in exactly 3 ways then the number of student - scholarship combinations is 10*3=30.
     There are other ways to visualize the problem, for example you could label the Y axis 1-10 and the X axis A,B,C and count all the grid points.
      I hope this helps, sorry about the confusion