30 students have entered a poetry competition. One student will win the Golden Prize and another student will win the Silver
Prize. Part a) How many different possible combinations of Golden Prize and Silver Prize
winners are there in total?
There is a second part to this question, which we will look at later. In order to solve this problem, we’re looking for the different combinations of
Golden Prize and Silver Prize winners. As there’re 30 students in total, the probability or chance of winning the Golden
Prize is one out of 30. This means that there are 30 possible Golden Prize winners.
Once the Golden Prize winner has been awarded, there is a one out of 29 chance for
the remaining students of winning the Silver Prize. This means that for each of the Golden Prize winners, there are 29 possible Silver
We can show this by firstly numbering each of the students from one to 30. If student number one wins the Golden Prize, then any of the other 29 students could
win the Silver Prize. It could be student number two, number three, number four, and on so on all the way
to number 30. This gives us 29 combinations.
If student number two won the Golden Prize, then student number one, student number
three, four, five, and so on could win the Silver Prize. Once again, this gives us 29 combinations.
We could repeat this with student number three, four, five, and so on winning the
Golden Prize up to student number 30. This means that for each of the 30 possible Golden Prize winners, there’re 29
possible Silver Prize winners. We need to multiply 30 by 29. This is equal to 870 as three multiplied by 29 is 87. There were, therefore, 870 different possible combinations of Golden Prize and Silver
The second part of the question says the following.
For a game, two opaque bags containing coloured balls are used. In the red bag, the 14 balls are red and numbered from one to 14. In the yellow bag, the 11 balls are yellow and numbered from one to 11. The game consists of picking a ball at random from one of the bags and then another
ball from either the same bag or the other bag. Each round of the game results in a list of two colour-number pairs. Part b) How many different lists can be created with this game?
The key start point here is that we are selecting two balls. We could select two red balls, a red ball followed by a yellow ball, a yellow ball
followed by a red ball, or two yellow balls. In total, we’ve 14 red balls and 11 yellow balls.
The probability of picking a specific red ball would be one out of 14. And picking a different specific red ball on our second pick would be one out of
13. This means to calculate the number of different combinations of two red balls. We need to multiply 14 by 13. This is equal to 182.
Next, we will look at a red ball followed by yellow ball. Once again, the probability of picking a specific red ball is one out of 14 and the
probability of picking a specific yellow ball is one out of 11. The total number of combinations is then calculated by multiplying 14 by 11. This is equal to 154.
Our third option was a yellow followed by a red. This is calculated in the same way: one out of 11 and one out of 14. 11 multiplied by 14 is also 154.
Finally, we need to consider two yellow balls. Picking two balls from this bag would be one out of 11 and one out of 10. This means that the number of different combinations of two yellow balls is 110 as 11
multiplied by 10 is 110. Adding these four numbers gives us 600.
Therefore, we can say that there are 600 different lists that can be created with
this game. In other words, there are 600 different combinations of balls that can be