-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Permutations and Combinations question (1 Viewer)
- Thread starter apollo1
- Start date
michaeljennings
Active Member
i think from memory it was 5x5x4!6 Envelopes are arranged in a straight line and numbered 1 to 6. Find the number of ways in which 6 different letters can be arranged one in each envelope if:
(i) letter A is not in envelope 1 and letter B is not in envelope 2.
michaeljennings
Active Member
dont lie to me brother =) I did this question yesterday
yeh stupid perm questions...first look at the case where there are no restrictions: 6!
next look at the case when A is in env 1 and B in env 2 (the only case we dont want), this occurs in 4! ways
so total ways therefore where A not in env1 and B not in env 2 is: 6! - 4! = 696
next look at the case when A is in env 1 and B in env 2 (the only case we dont want), this occurs in 4! ways
so total ways therefore where A not in env1 and B not in env 2 is: 6! - 4! = 696
michaeljennings
Active Member
Whattt I dont understand, I did this question yesterday and I swear the solution had the same reasoning you had to begin with and not this one
Because what about if A is in env1 and B is not in env2, you cant have this scenario either so you must subtract this one from total same as A is not in env1 but B is in env2
well if above soln is wrong this is the only other explanation
michaeljennings
Active Member
I editted my post can you take a look at it because now im confused
it's cause if it is A has 5 to choose from and B has 5 to choose from then they can coincide in one env, which is not allowed, only one letter is allowed in each env
_ _ _ _ _ _
6 spaces to be filled
if first is
B; 1 x 5!
c; 1 x 4 x 4!
D; 1 x 4 x 4!
etc until F
then add the no of ways = 5! +4(4x4!)
With my above reasoning i forgot to think of all cases:
So no restrictions means there are 6!=720 ways..
Now we dont want A is position one, A can be in position 1 in 5!=120 ways
Also, we dont want B in position two, B can be in position two in 5!=120 ways
Now, the number of ways A and B are not in 1st and 2nd position is 720 -120-120 = 480
However, when i minus by both i am minusing their intersection in 1st and 2nd position twice.
They are, as mentioned above, in 1st and 2nd position 4!=24 ways. Therefore i need to add this to 480 to get the total number of ways A and B are not in 1st and 2nd position.
Hence, the total number of ways is 504
So no restrictions means there are 6!=720 ways..
Now we dont want A is position one, A can be in position 1 in 5!=120 ways
Also, we dont want B in position two, B can be in position two in 5!=120 ways
Now, the number of ways A and B are not in 1st and 2nd position is 720 -120-120 = 480
However, when i minus by both i am minusing their intersection in 1st and 2nd position twice.
They are, as mentioned above, in 1st and 2nd position 4!=24 ways. Therefore i need to add this to 480 to get the total number of ways A and B are not in 1st and 2nd position.
Hence, the total number of ways is 504
michaeljennings
Active Member
Sorry if this sounds dumb but could you explain why is 'position 1'
the env's are arranged in a line 1-6, by pos 1 i am reffering to env 1, which is what we dont want A to be in
michaeljennings
Active Member
ohhh right makes sense sorry!