a) 20·9! = 5·4·9·8·7·6·5·4·3·2·1
a) 6P4 = 6! / (6-4)! = 6! / 2! = (6·5·4·3·2·1) / (2·1) = 6·5·4·3 = 360
a) 21! / (17!4!) = (21·20·19·18) / (4·3·2·1) = 7·5·19·9
Josh F.
tutor
Yes. 20x19! would be what you wrote. Your teacher is probably looking for you to write that as 20x19! = 20!
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05/03/21
Josh F.
tutor
For c) , what happens is that when you have a bigger integer factorial divided by a smaller integer factorial, the whole denominator factorial will cancel out with a part of the numerator factorial: 5! / 3! = (5x4x3x2x1) / (3x2x1) = 5x4 = 20.
So in c, the whole 17! canceled with most of the numerator factorial, which is why the numerator stops at 18.
Btw, that type of expression is exactly the kind of answer you get when you count permutations: How many ways can you choose a prez, vp, and treasurer from a group of 20?
Answer: 20P3 = 20! / (17!) = 20x19x18.
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05/03/21
Ally J.
Thank you so much, really appreciate your help! But the only thing I’m confused about at the moment is for a) like how would you get it in factorial form? Like for example 30x4!, so 10x3x4x3x2x1, The answer would is supposed to be 6!, how would you turn it into that is what I’m confused about.
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05/04/21
Josh F.
tutor
So if you are expected to turn it into a factorial expression, look for factors of the number multiplying the factorial that could add to the factorial. For instance, you are correct that 30x4! = 3x10x4!, however, if we want to expand 4!, the next number we need is 5. So, we recognize that 30=6x5. Then we get 6x5x4! = 6!
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05/04/21
Ally J.
Sorry I realised for a) I meant to write 20x19!, would it be 5x4x19x18x17....x1? I’m still confused. For c). numerator, why did u do that up till 18? Why not 17!? And why for only 4! ?05/03/21