Question

4 boys and 4 girls are arranged to sit in a row. Find the number of arrangements that can be made if :
(a) there is no restriction,
>>>(4+4)!=40320<<<
(b) the boys must sit separately,
>>>4!*4!=576<<<
(c) 2 girls must sit at the ends.
>>>(8-1)!*2!=10080<<<

is it correct?

Answer 1

@ganeshie8

Answer 2

a is correct
b think again
c wrong

Answer 3

for b :-
(b) the boys must sit separately,

tie all 4 boys in one bag
remaining 4 girls separate.

so total 1 bag + 4 girls = 5 objects can be permuted in 5! ways
boys inside bag can be permuted in 4! ways
so total arrangements = 5!*4!

Answer 4

oh i see...

Answer 5

how about part c then?

Answer 6

Errr, how can you tie all boys in one bag if all boys must sit separately in part b?

Answer 7

@ganeshie8

Answer 8

When boy must sit separately, girls must also sit separately, e.g.:
B G B G B G B G
Then 4!4! is correct for this case.
That's one arrangement. But you can also have a girl taking the first seat, i.e.
G B G B G B G B
So, you need to multiply your original answer by two.

Answer 9

When two girls must sit at two ends, you have:
\(G_1, x, x, x, x, x, x, G_2\)

So, you need to permute the 6 x's and the two G'.

Answer 10

i understand your point. but i am still confusing part b and c...

Answer 11

Which part(s) are you confused at?

Answer 12

b and c

Answer 13

I mean what..

Answer 14

why tie in a bag is wrong?

Answer 15

When you tie them together, they are connected and are no longer seperated, i.e.
You take A = B B B B, and you arrange
A G G G G

Answer 16

oh yes.

Answer 17

Clear?

Answer 18

clear. so what should we do in this case?

Answer 19

?

Answer 20

i mean, what should be the correct working step?

Answer 21

Read above comments

Answer 22

i cannot understand 6 x.

Answer 23

That's part c,right?

Answer 24

i got part b now. and go on to part c

Answer 25

Those x's are either boys or girls.

Answer 26

You only need to fix two girls (of course you need to permute that two girls as well) at the end, the others, you can mix and match, let the rest permute themselves.

Answer 27

when two girls sit in the two ends they can do it in 2 ways so remaining 4 boys and 2 girls = 6 students can sit in 6! = 720 ways
total no of ways = 2x720 = 1440

Answer 28

oh, i made it as (8-1)!*2! ...
understand now...i forgot 2 girls must sit at the ends....

Answer 29

:)