#Hello, can someone help solve and explain one of the Factorial problems.
plain smelt
grave trellisBOT
plain smelt
I'm mostly confused on the first one, I would be grateful if you guys can help.
tribal torrent
- for the just do it by writing how many options there are for each space
- treat the science posters as one object
- since it how many AREN'T, it will be the complement
stuck kestrel
I'm mostly confused on the first one, I would be grateful if you guys can help.
Since the math posters must be together at the ends (i.e., first and second place from an end), you can place them at either end.
First, place the math posters together at the left end and then find the possible number of arrangements of this case. Then, place them at the right end and do the same.
It is clear that whether you place them at the left end or the right, the number of arrangements will be the same. So just find the number of arrangements for one end and multiply by 2.
plain smelt
So, will it be 4 x 4! on (a)?
tribal torrent
no, it will be 2 x ... x 1 for math posters, because there is only two options for the first spot (one of the two maths posters), and one for the last one, as there is one maths poster left, and that is the only thing that can go on the end
in case you are confused "..." is stuff other than the maths posters in the above message
plain smelt
So it's 2 x 4! = 48?
plain smelt
Ah, I get it much clearer now, thank you very much.
tribal torrent
( 2 math ) * ( 4 not maths ) * ( 3 not math ) * ( 2 not math ) * ( 1 not math ) * ( 1 math )
ye
np
for b) treat all the science posters as one as one object, but dont forget to account for their rearrangements
plain smelt
Then it's 6 x 4 x 3!?
spiral vessel
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