gabriel wrote:Well, not entirely sure but how about 6c4*5!*4!= 43,200 .. If it is right I will give my explanation for it.

Regards

Correcto !

Here is my explanation..

_ B _ B _B _B _B _

The line would look like above after the boys are positioned. As you can see there are 6 empty spaces. Any 4 of those can be occupied by girls. This arrangement will make sure no two girls are next to each other.

# ways to arrange boys: 5!
# of ways to select 4 out of 6 positions for girls 6 C 4
# of ways to arrange girls in these 4 positions: 4!

Total = 5! * (6C4) * 4! = 43,200

Thats how you go about such P&C questions !

TkNeo: Excellent question. Thanks for letting us know this.

It is more like a game than a test question.

hi i am new to the form and i am student and the answer is 1152.let me know is this correct.if yes i will give the explanation.thank you

this problem seems like another one but the steps to find the answers are different. the other pblem is :
There are 6 guys and 4 girls.what is the probability of seating all the guys and girls in such a way that no two girls sit together?
the answer was
_b_b_b_b_b_b_

We can see that we have 6 boys to arrange, so there are 6! possible sequences for them.

There are 7 possible gaps and 4 girls to place in those spots. When we're not filling all the spots, we use the regular permutations formula:

nPk = n!/(n-k)!

in which n is the total number of objects and k is the number that we're using.

Here, n=7 and k=4, so:

7P4 = 7!/(7-4)! = 7!/3! = 7*6*5*4

6!*7*6*5*4 total arrangements that fit our desired criteria



Don t we have to do the same in this problem
then we will have
n=6 k=4
5!* 6!/(6-4)! ??????????????????????????????????????????

stufigol wrote:this problem seems like another one but the steps to find the answers are different. the other pblem is :

Stuart Kovinsky wrote:There are 6 guys and 4 girls.what is the probability of seating all the guys and girls in such a way that no two girls sit together?
the answer was
_b_b_b_b_b_b_

We can see that we have 6 boys to arrange, so there are 6! possible sequences for them.

There are 7 possible gaps and 4 girls to place in those spots. When we're not filling all the spots, we use the regular permutations formula:

nPk = n!/(n-k)!

in which n is the total number of objects and k is the number that we're using.

Here, n=7 and k=4, so:

7P4 = 7!/(7-4)! = 7!/3! = 7*6*5*4

6!*7*6*5*4 total arrangements that fit our desired criteria.

Don t we have to do the same in this problem
then we will have
n=6 k=4
5!* 6!/(6-4)! ??????????????????????????????????????????

That first part of your reply is my post from the problem with 6 boys and 4 girls; you're correct, we can solve this one the exact same way.

Here we have 5 boys and 4 girls, so there are 5! ways to arrange the boys.

Then we look at the gaps:

_b_b_b_b_b_

and see there are 6 "legal" spots in which to put girls. So, there are 6P4 ways to arrange the girls, giving us the 6!/(6-4)! in the second part of your product.

So, the answer is indeed:

5!*6!/2!

= 5*4*3*2*6*5*4*3*2/2

= 120 * 360

= 43200

TkNeo wrote:5 boys and 4 girls have to stand in a line such that no two girls are next to each other. how many possible ways ?

Let B denote a boy and G denote a girl.
Let us consider that 5 boys are standing in 5 slots. [ B1 B2 B3 B4 B5 ]
Now,we have 6 empty slots. [ | B1 | B2 | B3 | B4 | B5 | ]

Since, no two girls are next to each other , so selecting any 4 of the 6 slots . 6C4 x 4! {x4! since the girls can also exchange places between them }

Also,the boys can also exchange places between them.Hence,another multiplication factor of 5!.

Hence,the answer will be 6C4 x 4! x 5! = 15 x 24 x 120 = 43200

I thought the question was self sufficient. Stuart, am i overseeing something ?

Based on what i interpret....

5 Boys have 6 empty spots around them....

_B_B_B_B_B_

Let the girls pick four of the empty sport....

i.e. 6P4 = 360

The 5 boys can be adjusted in 5! ways

So total combinations = 5! * 360 = 120 * 360 = 43200 ways.

See this is a very easy question related to permutation and combination :
It can be easily solved by following procedure..
first read the question carefully.. There are 5 boys and 6 girls...and no two girls can line up next two each other.
Let us represent boys by B and girls by G. So for no two girls to line up next to each other, there must be atleast one boy between two girls.

so we first arrange boys and create empty spaces between them where girls can be placed. let this empty spaces being symbolized by X.

X B X B X B X B X B X

so we can see there are 6 empty spaces where these girls could be placed.

No. of ways in which 4 spaces out 6 can be selected is given by 6C4 = 6!/2!/4! = 6x5/2=15
In each of the above case, boys can be arranged in their 5 places in 5! = 120 ways
& girls can be arranged in their 4 places in 4! = 24 ways

so total no. of ways in which boys and girls can be lined up = 15 x 120 x 24 = 43200 ways.

Btw girls are very talkative and they must not be lined next to each other..otherwise you will get a fish market effect..so keep calculating the permutations and combinations..

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