Permutations and Combinations question (1 Viewer)

[apollo1]

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]

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.

i think from memory it was 5x5x4!

 

[apollo1]

ur both wrong.

 

[michaeljennings]

ur both wrong.

dont lie to me brother =) I did this question yesterday

 

[math man]

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

 

[michaeljennings]

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

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

 

[math man]

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

well if above soln is wrong this is the only other explanation

 

[michaeljennings]

I editted my post can you take a look at it because now im confused

 

[apollo1]

the solution I have says its 504.

 

[math man]

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

 

[hup]

is the answer 504?

 

[apollo1]

is the answer 504?

yes how did you get it?

 

[hup]

yes how did you get it?

_ _ _ _ _ _

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!)

 

[math man]

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

 

[math man]

_ _ _ _ _ _

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!)

This is perfectly right, however in comb/perm questions you dont wanna be looking at every case, you wanna look at total cases and then minus number of cases in which our restriction is breached, but props to your soln

 

[michaeljennings]

=

Now we dont want A is position one, A can be in position 1 in 5!=120 ways

Sorry if this sounds dumb but could you explain why is 'position 1'

 

[math man]

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]

ohhh right makes sense sorry!

 

[hup]

because youre on this

denotes the number of ways in which 2n distinct items can be paired off (order unimportant)

obv =1

explain why =3

and deduce that

 

[td547]

_ _ _ _ _ _

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!)

wow hup!! great work mate, solution is perfect, tyy!