Help with Permutations and Combinations (1 Viewer)

[moderntortoisecat]

Find the number of possible arrangements of the letters in the word PENCILS if
a) E comes before I
B) There are three letters between E and I

I have seen this question on a thread a long time ago but for a), im confused with how to know that there are equal possibilities of e coming before i and i coming before e. Any explanations would be really appreciated

 

[liamkk112]

Find the number of possible arrangements of the letters in the word PENCILS if
a) E comes before I
B) There are three letters between E and I

I have seen this question on a thread a long time ago but for a), im confused with how to know that there are equal possibilities of e coming before i and i coming before e. Any explanations would be really appreciated <3

a) the simplest, but slowest method is by cases:
lets think of 7 blank spaces: _ _ _ _ _ _ _
for E to come before I, I can take all spaces except the leftmost one.
for the case where I is in the rightmost space (_ _ _ _ _ _ I), we have 6! ways to arrange the other letters
for the case where I is in the second rightmost space (_ _ _ _ _ I _), we have 5 choices to arrange all letters but E, and then 5! ways to arrange the rest
where I is in the third rightmost space (_ _ _ _ I _ _), we have 5x4 choices to arrange all letters but E, and then 4! ways to arrange the rest
where I is in the fourth rightmost space (_ _ _ I _ _ _), we have 5x4x3 choices to arrange all letters but E, and then 3! ways to arrange the rest
fifth rightmost (_ _ I _ _ _ _), we have 5x4x3x2 choices to arrange all letters but E, and then 2! ways to arrange the rest
sixth rightmost (_ I _ _ _ _ _),, we have 5x4x3x2x1 choices to arrange all letters but E, and then 1! ways to arrange the rest

summing these possibilities up we get that there are 2520 possibilities

there is probably a more nuanced method but since the word is small this is possible to do

b) first a more nuanced method:
lets place E and I with 3 letters inbetween, like this: E _ _ _ I
notice now E and I are fixed. we now have 5 letters remaining.
to fill in these 3 letters, we have 5x4x3 choices. lets now imagine 3 blank spaces: _ _ _
we can fill in the blank spaces with 3 groups: the group consisting of E _ _ _ I, and the other 2 groups being the leftover letters
now there are 3! different permutations of these groups
hence in total there are 3! x 5x4x3 = 360 ways to do this

we can also do by cases, by noting there are 3 different ways to arrange the situation:
_ E _ _ _ I _ or _ _ E _ _ _ I or E _ _ _ I _ _
in all these cases there are 5! ways to arrange the remaining letters
so we have 5! + 5! + 5! total ways or 360 ways to do this

 

[Luukas.2]

Find the number of possible arrangements of the letters in the word PENCILS if
a) E comes before I
B) There are three letters between E and I

I have seen this question on a thread a long time ago but for a), im confused with how to know that there are equal possibilities of e coming before i and i coming before e. Any explanations would be really appreciated <3

For every permutation in which the E comes before the I, swapping these letters produces a corresponding permutation where the I comes before the E. It is impossible for one of them not to precede the other (as they can't both be in the same position), so the set of all possible permutations can be divided into two equally sized and non-overlapping subsets, one with E before I and the other with I before E. Thus:

# E before I permutations = # I before E permutations​

and

# E before I permutations + # I before E permutations = total number of permutations = 7!​

and so

# E before I permutations = 7! / 2 = 2520 permutations
​

Modifying @liamkk112's counting method slightly:

• E - - - - - - cases: Position E in 1 way, I in 6 ways, rest in 5! ways ---> 6 x 5!​
• - E - - - - - cases: Position E in 1 way, I in 5 ways, rest in 5! ways ---> 5 x 5!
• - - E - - - - cases: Position E in 1 way, I in 4 ways, rest in 5! ways ---> 4 x 5!
• - - - E - - - cases: Position E in 1 way, I in 3 ways, rest in 5! ways ---> 3 x 5!
• - - - - E - - cases: Position E in 1 way, I in 2 ways, rest in 5! ways ---> 2 x 5!
• - - - - - E - cases: Position E in 1 way, I in 1 way, rest in 5! ways ---> 1 x 5!
• - - - - - - E cases: Position E in 1 way, I in 0 ways, so not possible

# E before I permutations = (6 + 5 + 4 + 3 + 2 + 1) x 5! = 21 x 5! = 7 x 3 x 5! = 7 x (6 / 2) x 5! = 7! / 2 = 2520 permutations

@liamkk112's answer to (b) has a mistake in it, but the method is sound. The answer is 720 permutations.

 

[ISAM77]

I did part a) the same way as Lukkas above.

b)
6 possible ways to set out E and I:

E _ _ _ I _ _
_ E _ _ _ I _
_ _ E _ _ _ I

Note that E and I can be written in reverse order so that there are 3x2 = 6 possible ways. The remaining three ways are also shown below:

I _ _ _ E _ _
_ I _ _ _ E _
_ _ I _ _ _ E

...

Now, there are 5 remaining gaps for 5 different letters. This can be filled out 5! number of ways.
Hence, 6 * 5! = 720 permutations

 

[liamkk112]

I did part a) the same way as Lukkas above.

b)
6 possible ways to set out E and I:

E _ _ _ I _ _
_ E _ _ _ I _
_ _ E _ _ _ I

Note that E and I can be written in reverse order so that there are 3x2 = 6 possible ways. The remaining three ways are also shown below:

I _ _ _ E _ _
_ I _ _ _ E _
_ _ I _ _ _ E

...

Now, there are 5 remaining gaps for 5 different letters. This can be filled out 5! number of ways.
Hence, 6 * 5! = 720 permutations

oh shoot i forgot that I can go before E lol, i think i forgot because of part a

 

[Luukas.2]

Follow Up... how many permutations of the word PENCILS will have the "I", "C", and "E" occur in that order?