Suppose a city with Three type of coins ?! in a city we have tree type 1 dollar, 2 dollar, 3 dollar of coins. we want to pay for a 20 dollar product. how many ways we can pay for a 20 dollar product, if the seller has no money and number of 1 dollar coin is more than 2 dollar coin. 

1) 20
2) 21
3) 38
4) 40

This is a 2010 contest question that the answer sheet say (2) is True, but there is no way to reach it ! any hint or idea ?
 A: 1 way with 6x \$3 coins  (\$2 with \$1 and \$2, so must be 0 \$2 coins)
2 ways with 5x \$3 coins (\$5 with \$1 and \$2, so between 0 and 1 \$2 coins)
3 ways with 4x \$3 coins (\$8 with \$1 and \$2, so between 0 and 2 \$2 coins)
4 ways with 3x \$3 coins (\$11 with \$1 and \$2, so between 0 and 3 \$2 coins)
5 ways with 2x \$3 coins (\$14 with \$1 and \$2, so between 0 and 4 \$2 coins)
6 ways with 1x \$3 coins (\$17 with \$1 and \$2, so between 0 and 5 \$2 coins)
7 ways with 0x \$3 coins (\$20 with \$1 and \$2, so between 0 and 6 \$2 coins)
Makes a total of 28 ways. Either there are more conditions that you haven't stated, or the question is wrong.
A: Discrete positive solution for the plane x+2y+3z=20 where x>y always. The count comes to be 28. 
A: Since we need more 1-dollar coins than 2-dollar coins in the payment, we can investigate solutions using only 1s and 3s, then split the 3s into 2+1 for more solutions. 
Counts will be given as $(x,y,z)$ where $x+2y+3z=20$.
So $(11,0,3)$ splits into 3 additional solutions that use \$2 coins: $(12,1,2), (13,2,1), (14,3,0)$ - and in general if we start with $n$ \$3 coins (and no \$2 coins) we generate another $n$ solutions by splitting the \$3s.
So the total solutions arising from having $6,5,4,3,2,1,0$ \$3s and no \$2s are $7+6+5+4+3+2+1 = 28$.
$7$ of these solutions have no \$2-dollar coins. If this is forbidden, you would reach $21$ options.
