# Lottery exam question

I am studying a past examination paper however I do not have the answers to this paper so I can't check if I am right or wrong. I have just reached a section on combinatorics and counting and I have come across a few troubling questions.

The lottery rules state that you choose five different numbers in the range 1 to 53 and write them in an order of your choice on an entry form.

Question a) asks : On a lottery day, a machine selects ten balls from a sphere containing 53 balls numbered 1 to 53 and arranges them in a row. How many different outcomes can this process have? (write answer to 2 significant figures in scientific notation)

My answer to this was $53 \choose 10$ which is $1.9 \times 10^{10}$

I used choose (binomial coefficient) because the order doesnt matter and there not repetitions.

part b) to this question however, Im not sure how to calculate:

Part (b) asks : You win prize 1 if the first five balls selected by the machine match the five numbers you chose, and are arranged in the order you wrote them on your entry form. In how many of the outcomes of part (a) do you win prize 1? (You can write your answer in scientific notification and to 2 significant figures)'

I started by calculating the number of permutations you can actually have of choosing 5 lottery numbers which was 53!. My problem from here is that I don't know where else to start calculating and I'm really not sure if I have even started to calculate this in the right way? Please could someone explain to me the best way of doing this in as much detail as they can so that I can learn from this?

In part (a) the order does matter: this is indicated by the statement that the balls are arranged in a row. If they’re in a row, they automatically have an order. Thus, the correct answer is

$$53\cdot52\cdot\ldots\cdot44=\frac{53!}{43!}=\binom{53}{10}\cdot 10!\approx 7.1\times 10^{16}\;.$$

For (b), we’ll count the winning outcomes. There is only one winning choice for the first ball. In fact, there’s only one winning choice for each of the first five balls. The sixth ball drawn can be any of the remaining $53-5=48$ balls, the seventh can be any of the $47$ balls still left, and so on, so there are

$$1^5\cdot48\cdot47\cdot46\cdot45\cdot44=205,476,480$$

winning outcomes.

• I'm not sure if you understood what part b was asking? part b asks if the 'first 5 balls' match the 5 balls you chose, correct me if I am wrong but you are looking at the last 5 balls and working out the winning outcomes from them? Aug 22, 2016 at 18:03
• @RJB: Yes, I understood (b). In order for the first $5$ balls to match the ones that you chose, they must be precisely those $5$ balls: there is only one first ball that can give you a winning draw, only one second ball, and so on. The last $5$ balls drawn, however, can be any of the remaining $48$ balls in any order. This is exactly the calculation that I made. Aug 22, 2016 at 18:06
• ahh okay I see what you are saying now and yes you are correct, thanks for your help! Aug 22, 2016 at 18:11
• @RJB: You’re welcome! Aug 22, 2016 at 18:11
• Hi Brian, I am answering the following question which is part c) this states "You win Prize 2 if the first five balls selected by the machine match the five numbers you chose, and are arranged in any order (including that for which you win Prize 1). In how many of the outcomes of part (a) do you win Prize 2? (You may write your answer in scientific notation, to 2 significant figures.)" I tried working this out by gathering that the first choice would have 53 winning outcomes, the second would have 52, third would have 51 and so on until the 5th choice. This gave me 344362200 but this - Aug 24, 2016 at 14:08