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# factorials watch

1. Can anyone explain factorials for AS stats/core 2 AQA. I sometimes make a mistake in probability questions when deciding how many different options/arrangements there are. Is there an easy way of doing this?
EG) On three particular days, the ferry departs from port D on time.
Find the probability that, on these three days, the ferry arrives at port A early once, on time once and late once.
(QUESTION 3 ON http://filestore.aqa.org.uk/subjects...P-QP-JUN15.PDF).

for this question, I multiplied the three probabilities by 3, rather than 3! (or 6). After seeing the mark scheme, I understood why there were 6 arrangements (EG. LEO, ELO, EOL, OEL, LOE, OLE) however, is there a way in the exam I can do this without fail, and more quickly.

2. (Original post by a.myhall)
Can anyone explain factorials for AS stats/core 2 AQA. I sometimes make a mistake in probability questions when deciding how many different options/arrangements there are. Is there an easy way of doing this?
EG) On three particular days, the ferry departs from port D on time.
Find the probability that, on these three days, the ferry arrives at port A early once, on time once and late once.
(QUESTION 3 ON http://filestore.aqa.org.uk/subjects...P-QP-JUN15.PDF).

for this question, I multiplied the three probabilities by 3, rather than 3! (or 6). After seeing the mark scheme, I understood why there were 6 arrangements (EG. LEO, ELO, EOL, OEL, LOE, OLE) however, is there a way in the exam I can do this without fail, and more quickly.

Presumably when you multiplied by 3, you were thinking of the situation where you have two the same and one different - at a guess.

The number of arrangements n objects where you have p the same of one type is n!/p!
So, 3 objects and 2 the same gives 3!/2! = 3 arrangements.

If you have p of one type and q of another type, it would be n!/(p!q!),
(and if these are the only two types, p+q=n, and you have the standard binomial coefficient)
Etc.

So, always start with n!, and then divide by the factorial of the number of each identical type.
3. (Original post by ghostwalker)
Presumably when you multiplied by 3, you were thinking of the situation where you have two the same and one different - at a guess.

The number of arrangements n objects where you have p the same of one type is n!/p!
So, 3 objects and 2 the same gives 3!/2! = 3 arrangements.

If you have p of one type and q of another type, it would be n!/(p!q!),
(and if these are the only two types, p+q=n, and you have the standard binomial coefficient)
Etc.

So, always start with n!, and then divide by the factorial of the number of each identical type.
so 3!/1! in my example. Because there are 3 different options (early, on time, or late) and we're looking for a situation where they all occur once?
4. (Original post by a.myhall)
so 3!/1! in my example. Because there are 3 different options (early, on time, or late) and we're looking for a situation where they all occur once?
Since they're all different, you could just leave it as 3! You don't need to be concerned further about types only occuring once.

BUT, if you're going to take the repetitions into account, then there is one E, one T, and one L, so

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