Possibilities are Student A has 2 and the others 1, Student B has 2 and the others 1, Student C has 2 and the others 1. So let's just find out how many possibilities lie in each of these three scenarios, and then multiply by three.
Student A may have any two books of four (order is irrelevant), giving six possibilities (1&2, 1&3, 1&4, 2&3, 2&4, 3&4). Of each of these six possibilities, there are only two ways to distribute the remaining two books to each student. So, there are twelve possibilities for each of the three listed scenarios.
12 x 3 = 36. C.
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Ist Person ---> 4 ways
2nd Person ---> 3 ways
3 rd Person ---> 2 ways
Total = 4*3*2 = 24 ways Hence D.
2nd Person ---> 3 ways
3 rd Person ---> 2 ways
Total = 4*3*2 = 24 ways Hence D.
This does not take into account that one person may have two books, giving that slot six possibilities. See above explanation. This answer would be correct if each student could only have one book.bjp2008 wrote:Ist Person ---> 4 ways
2nd Person ---> 3 ways
3 rd Person ---> 2 ways
Total = 4*3*2 = 24 ways Hence D.
Last edited by Feep on Sat Apr 11, 2009 2:13 am, edited 1 time in total.
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concentrate on the word atleast...
out of the three one person will get 2 books and the other two will get 1 book each.
consider the first student gets two books..
this can be done in 4c2x2x1 = 12 ways.
now similarly if we consider the second person to get 2 books we have another 12 ways and for the third person to get 2 books we have another 12 ways...
thus 12x3= 36
Ans C
out of the three one person will get 2 books and the other two will get 1 book each.
consider the first student gets two books..
this can be done in 4c2x2x1 = 12 ways.
now similarly if we consider the second person to get 2 books we have another 12 ways and for the third person to get 2 books we have another 12 ways...
thus 12x3= 36
Ans C
