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.