1. Arun and Disha have some numbered discs to share out between them. They want to end up with one pile each, not necessarily of the same size, where Arun's pile contains exactly one disc numbered with a multiple of 2 and Disha's pile contains exactly one disc numbered with a multiple of 3 . For each case below, either count the number of ways of sharing the discs, or explain why it is impossible to share them in this way.
(a) They start with ten discs numbered from 1 to 10 .
(b) They start with twenty discs numbered from 1 to 20 .
SolUtion
(a) Consider the disc numbered 6. Assume for the moment that it is in Arun's pile. Then all the other discs numbered with a multiple of 2 must be in Disha's pile.
Disha's pile must also contain one of the two remaining discs numbered with a multiple of 3 (with the other in Arun's pile), so there are ${ }^{2} C_{1}=2$ ways to arrange the remaining discs numbered with a multiple of 3 .
Once the discs containing multiples of 3 have been allocated to a pile, there are three remaining discs (those numbered 1,5 and 7), each of which could go in either of the two piles, so there are $2^{3}=8$ ways this can be done.
So there are $2 \times 8=16$ ways of distributing the discs if the disc numbered 6 is in Arun's pile.
Now assume that the disc numbered 6 is in Disha's pile. Then all the other discs numbered with a multiple of 3 must be in Arun's pile.
Arun's pile must also contain one of the four remaining discs numbered with a multiple of 2 (with the other three all in Disha's pile), so there are ${ }^{4} C_{1}=4$ ways to arrange the remaining discs numbered with a multiple of 2 .
Once the discs containing multiples of 2 have been allocated to a pile, there are again three remaining discs ( 1,5 and 7 ), and as before there are 8 ways these can be allocated to the piles.
So there are $4 \times 8=32$ ways of distributing the discs if the disc numbered 6 is in Disha's pile.
Hence there are $16+32=48$ ways they can share the discs out.
(b) Consider the three discs numbered 6,12 and 18, which are multiples of both 2 and 3 . Since there are three of them, when the discs are shared out, one of Arun's and Disha's pile must contain (at least) two of them. But that is against the rules for how Arun and Disha want to share the discs out, so it is impossible to share them out.