Perms and Combs question - Help! (1 Viewer)

[EpikHigh]

Can anyone help me out with this question and give a nice explanation please I would be so greatful and will +rep, its from the Year 12 Cambridge book Ex 10G Q25 a-d

 

[bleakarcher]

your class already started perms and combs?

 

[EpikHigh]

Yeah, really interesting stuff haha, but this question has me a bit stumped, we've practically finished perms and combs.

 

[RealiseSomethin]

yea my school started it awhile ago as well

 

[RealiseNothing]

You use the pattern "1+2+3+...+n".

For a) "n" is 20, so it's 1+2+3+...+20 which is 210.

For b) split it up into 2 parts, the amount of routes from A to C multiplied by the amount of routes from C to B.

So from A to C, let "n" be 3, so 1+2+3 = 6

From C to B, let "n" be 5, so 1+2+3+4+5 = 15

Hence 6 times 15 = 90 possible routes from A to B through C.

Will do others now.

 

[RealiseNothing]

For c) we must not include the top row.

So lets take the total amount of routes and subtract the routes that go along the top row.

The routes that go along the top row and go to atleast the 2nd line from the right have 8 routes. So 210-8 = 202.

The routes that go to the 3rd line from the right have n=6 so 1+2+3+4+5+6 = 21 routes. Thus 202-21 = 181.

The routes that go to the 4th line from the right have n=10 so 1+2+3+...+9+10 = 55 routes. Hence 181-55 = 126.

So there are 126 routes if you don't go along the top row.

 

[RealiseNothing]

d) is the same as c), so 126 routes