A machine dispenses four kinds of tokens. Red tokens are worth \(25\) points, blue tokens are worth \(10\) points, green tokens are worth \(5\) points, and yellow tokens are worth \(1\) point. The machine never runs out of tokens to dispense.
To get tokens from the machine, people enter a total number of points, and the machine will dispense the fewest number of tokens that total that number of points. For example, if someone enters \(30\) points, the machine will dispense \(1\) red token and \(1\) green token, because their total value is \(25+5=30\) points, and this total cannot be obtained using fewer tokens.
For how many different total numbers of points will the machine dispense exactly \(4\) tokens?
We start by noting how many of the \(4\) tokens can be a particular colour. There can be up to \(4\) red tokens because they have the largest value, so the machine cannot obtain the same total number of points using fewer tokens. There can also be up to \(4\) yellow tokens, because \(4\) yellow tokens have a total value of \(4\) points, which is less than the value of \(1\) green token. The number of blue tokens must be less than \(3\), because \(3\) blue tokens have a total value of \(30\) points, which can be obtained using fewer tokens (namely \(1\) red token and \(1\) green token). Similarly the number of green tokens must be less than \(2\), because \(2\) green tokens have a total value of \(10\) points, which is the value of \(1\) blue token. Finally, we cannot have \(2\) blue tokens and \(1\) green token, because they have a total value of \(25\) points, which is the value of \(1\) red token.
Now we count the number of combinations of \(4\) tokens that can be dispensed from the machine.
Case 1: There are \(4\) red tokens.
In this case, the only total number of points possible is \(100\).
Case 2: There are \(3\) red tokens.
If there are \(3\) red tokens, then the
other token could be blue, green, or yellow. The total numbers of points
would then be \(85\), \(80\), or \(76\), respectively. Thus, there are \(3\) different total numbers of
points.
Case 3: There are \(2\) red tokens.
If there are \(2\) red tokens, then the
other \(2\) tokens are either both the
same colour, or are two different colours.
If the other \(2\) tokens are both the same colour, then they can either be blue or yellow. The total numbers of points would then be \(70\) or \(52\), respectively. Note that they cannot be green because the number of green tokens must be less than \(2\).
If the other \(2\) tokens are two different colours, then they can be blue and green, blue and yellow, or green and yellow. The total numbers of points would then be \(65\), \(61\), or \(56\), respectively.
Thus, there are \(5\) different total numbers of points if there are \(2\) red tokens.
Case 4: There is \(1\) red token.
If there is \(1\) red token, then the
other \(3\) tokens are either all the
same colour, \(2\) are the same colour
and \(1\) is a different colour, or
they are all different colours.
If the other \(3\) tokens are all the same colour, then they must all be yellow. Therefore, the total number of points would be \(28\). Note that they cannot all be blue or all be green because the number of blue tokens must be less than \(3\) and the number of green tokens must be less than \(2\).
If \(2\) of the other \(3\) tokens are the same colour and \(1\) is a different colour, then there could be \(2\) blue and \(1\) yellow, \(1\) blue and \(2\) yellow, or \(1\) green and \(2\) yellow. The total numbers of points would then be \(46\), \(37\), or \(32\), respectively. Note that we already determined that we cannot have \(2\) blue tokens and \(1\) green token.
If the other \(3\) tokens are all different colours, then the total number of points would be \(41\).
Thus, there are \(5\) different total numbers of points if there is \(1\) red token.
Case 5: There are no red tokens.
If there are no red tokens, then there are either \(4\) tokens of one colour, \(3\) tokens of one colour and \(1\) token of another colour, \(2\) tokens of one colour and \(2\) tokens of another colour, or \(2\) tokens of one colour and \(1\) token of each of the other two
colours.
If there are \(4\) tokens of one colour, then they must be all yellow. Therefore, the total number of points would be \(4\).
If there are \(3\) tokens of one colour and \(1\) token of another colour, then there could be \(1\) blue token and \(3\) yellow tokens or \(1\) green token and \(3\) yellow tokens. The total numbers of points would then be \(13\) or \(8\), respectively.
If there are \(2\) tokens of one colour and \(2\) tokens of another colour, then there must be \(2\) blue tokens and \(2\) yellow tokens. Therefore, the total number of points would be \(22\).
If there are \(2\) tokens of one colour and \(1\) token of each of the other two colours, then there must be \(1\) blue token, \(1\) green token, and \(2\) yellow tokens. Therefore, the total number of points would be \(17\).
Thus, there are \(5\) different total numbers of points if there are no red tokens.
In total, there are \(1+3+5+5+5=19\) different total numbers of points for which the machine will dispense exactly \(4\) tokens.
The \(19\) possible totals obtained using exactly \(4\) tokens are as follows.
\(4\), \(8\), \(13\), \(17\), \(22\), \(28\), \(32\), \(37\), \(41\), \(46\), \(52\), \(56\), \(61\), \(65\), \(70\), \(76\), \(80\), \(85\), \(100\)