Hey i got the way how u also solved the sum . thanks! 
its not that difficult after all
its not that difficult after all
harsh.champ wrote:Ofcourse not,where is the question of any offence??bhumika.k.shah wrote:hey thanks i got wht u did. but i wanna know thing. hope u dont get offended .
is there any difference b/w the way u solved it and the way thephoenix solved it ?????
harsh.champ wrote:The italics text was given by rahul.s.bhumika.k.shah wrote:Hey ,
Could you elaborate further on how u got the highlighted part ?
harsh.champ wrote:Hey rahul.s,rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?
[spoiler]OA is 3/10 [/spoiler]
[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]
Did u delete ur post after posting for the 1st time??
As for the soln.,
A and B will star together.Now, the remaining 1 role can be selected in 3C1 ways.
Also,3!ways in which roles can be interchanged.
Total no. of ways = 5P3 = 60
Hence, 3x6/60 = 3/10 which is the OA.
I hope you can now see your mistake.
You dont have to consider separately as you did over here.
my solution: 1/5 * 1/4 = 1/20.
The bold-faced text by me.
Which part did you not understand??
Anyways I will explain how I did it.
3 roles(suppose R1 ,R2 and R3 ) are there.
If A and B star together , only 1 role is left .The 3rd actress can be selected among the remaining 3 left in 3C1 ways.
Now, they can also interchange the roles between them in 3! = 6 ways.
So,no. of ways = 6 x 3 =18 .
For total no. of ways ,we can select the 3 roles among 5 actresses in 5C3(5!/3! x 2!) ways.
Also they can interchange the roles betweeen them in 3! ways.
So, total no.. of ways = 5C3 x 3! = 60 ways . [5C3 x 3! = 5P3 = 5! / 2!]
So, the ans. is 18/60 = 3/10.
I hope my soln. approach is clear to you Bhumika or I can elaborate more if you want.
This community is there so that we can share our views,help each other and hope for getting 700+ scores in the exam.
So in case of any doubt,JUST BRING IT ON.
Now,as for thephoenix's method :-For objective ,it is okay and short but if you are considering the proper approach,you should also conside the permutations(The fact that the 3 roles can also be exchanged between them in 3 ways).
"thephoenix" has only considered the selections not the permutations.
In this question,the 3! cancels in the numerator as well as in the denominator ,so what you see is the final product 3/10.
I had shown the permutations,also so that any community member reading the solution understands how the approach is made.Hence,my method seems lengthier but if you get the knack of the approach you can do it in any manner.
Hope you understand the difference between the methods.Just read both the solutions once again and you will get it.:)
