a five member committee is to be formed from a group of five x and 9 y. if the committee must have at least 2 x and 2 y, in how many diff ways can the committee be chosen.
ans is 1200.
I get 3600.
5!/2!3! * 9/2!7! *10=3600
solution says 1200 as not 2 but 3 ! for one group so 1200 as we will have 3 possible arrangements from the groups we choose.
Thanks very much
combination/permutation-PR cat - hard one?
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Since atleast two members to be chosen from both the goups, therefore 2 combinations can be possible..............
1. choose 2 from group of 5 and 3 from group of 9
= 5C2 * 9C3 = 840
2. choose 3 from group of 5 and 2 from group of 9
= 5C3 * 9C2 = 360
total= 840+360=1200
1. choose 2 from group of 5 and 3 from group of 9
= 5C2 * 9C3 = 840
2. choose 3 from group of 5 and 2 from group of 9
= 5C3 * 9C2 = 360
total= 840+360=1200
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I think I get it.
For combination, whether we choose 3 or 2 from 5 it is still 10, so choosing 3 from 9 =10*84=840 and choosing 3 from 5 and 2 from 9 is 10*360= total of 360+840=1200. ANS.
Thanks
Khurram
For combination, whether we choose 3 or 2 from 5 it is still 10, so choosing 3 from 9 =10*84=840 and choosing 3 from 5 and 2 from 9 is 10*360= total of 360+840=1200. ANS.
Thanks
Khurram