How many different ways are there for 4 boys and 2 girls to sit in a row?

Let us consider the 4 girls as a single person so there are now 5 persons in total (4 boys + 4 girls as one group or one person). So the number of ways in which 5 persons can be arranged is 5! ways and also the 4 girls can sit among themselves in 4! way so the total number of ways of seating this group of 5 is 5!×4! ways .

So the total number of ways in which 4 boys and 4 girls can be seated so that not all the girls sit together is given by= (Total no. of ways in which all 8 can sit)-(No. of ways the group of 5 can sit where the 4 girls all together is taken as a single person)

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in how many ways can 4 boys and 2 girls be seated if
i)the 2 girls are to seat next to eachother

We have two cases:

1. Girl #1 sits left of Girl #2
2. Girl #2 sits left of Girl #1

In case 1, imagine that Girl #1's right wrist is handcuffed to Girl #2's left
wrist. Thinking of it that way, we actually only have 5 "things" to arrange,
namely 4 boys and 1 pair of handcuffed girls.

We can seat the first "thing" on the far left any of 5 ways.

For each of those 5 ways that we can seat a "thing" on the far left, we can
choose any of the remaining 4 things to sit second from the left.  That's 5x4
or 20 ways to seat something first and second from the left. 

For each of those 5x4 or 20 ways that we can seat two "things" on the far left,
we can choose any of the remaining 3 things to sit third from the left.  That's
5x4x3 or 60 ways to seat something first, second and third from the left.

For each of those 5x4x3 or 60 ways that we can seat three "things" on the left,
we can choose either of the remaining 2 "things" to sit fourth from the left.
That's 5x4x3x2 or 120 ways to seat something first, second, third and fourth
from the left.

For each of those 5x4x3x2 or 120 ways that we can seat four "things" on the
left, we can choose only the 1 remaining "thing" to sit fifth from the left, or
on the far right.  That's 5x4x3x2x1 or 120 ways to seat something first,
second, third, fourth and fifth from the left.  Now all are seated.

So there are 5x4x3x2x1 or 120 ways for case 1.

Now if we imagine handcuffing the two girls the other way, that is, imagine
that Girl #2's right wrist is handcuffed to Girl #1's left wrist.  Then we get
exactly the same number 120 again.

So putting the two 120's together, we get 240 ways of seating them so that the
two girls sit side by side. 

This can also be thought of as

1. There are 2P2 of 2! or 2x1 or 2 ways to handcuff the two girls.
2. For each of the 2P2 or 2! or 2 way to handcuff the two girls, there are 
   5P5 or 5! or 5x4x3x2x1 or 120 ways to seat the 5 "things".
3. So the result can be written as (2P2)(5P5) or (2!)(5!) or (2)(120) = 240.


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ii)the 2 girls are not to seat next to each other.

Here the strategy is 

1. First we determine the number of ways any of the 6 can sit anywhere.

then

2. Subtract the 240 ways from part (i), which will remove all cases where
the girls sit together.

First we calculate the number of ways any of the 6 can sit anywhere:

We can seat the first person on the far left any of 6 ways.

For each of those 6 ways that we can seat a person in the far left seat, we can
choose any of the remaining 5 people to sit in the second seat from the left. 
That's 6x5 or 30 ways to seat somebody in the first two seats on the left. 

For each of those 6x5 or 30 ways that we can seat somebody in the two leftmost
seats, we can choose any of the remaining 4 people to sit in the third seat
from the left.  That's 6x5x4 or 120 ways to seat somebody in the first three
seats on the left.

For each of those 6x5x4 or 120 ways that we can seat somebody in the three
leftmost seats, we can choose any of the remaining 3 people to sit in the
fourth seat from the left.  That's 6x5x4x3 or 360 ways to seat somebody in the
first four seats on the left.

For each of those 6x5x4x3 or 360 ways that we can seat somebody in the four
leftmost seats, we can choose any of the remaining 2 people to sit in the fifth
seat from the left.  That's 6x5x4x3x2 or 720 ways to seat somebody in the first
five seats on the left.

For each of those 6x5x4x3x2 or 720 ways that we can seat somebody in the fifth
leftmost seats, we can only choose the one remaining person to sit in the sixth
seat from the left, or the last seat on the right.  That's 6x5x4x3x2x1 or 720
ways to seat somebody in all 6 seats in every possible order.

So the final answer will be these 720 MINUS the 240 ways from part (i) where 
the two girls sit together, and we get 720 - 240 or 480 ways when the girls
do not sit together.

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The results of this problem can also be thought of and written either in terms
of permutations or factorials:

We can arrange the 6 people 6P6 or 6! or 720 ways.

We can "handcuff" the two girls 2! or 2 ways.

We can seat the 5 things (4 boys and a pair of handcuff girls) 5P5 or 5! or 120 ways.

So, if you prefer, in terms of permutations or factorials, we have:

6P6 - (2P2)(5P5) = 6! - (2!)(5!) = 720 - 2(120) = 720 - 240 = 480. 

Edwin