[shanks4mba]

Originally Posted by gunner4life

1. A train running between A and B has 9 stops in between. 7 passengers enter the train somewhere in between A and B There are no restrictions on no. of tickets that can be purchased on a station. In how many ways can the tickets be purchased?

2. From 4 gents, and 4 ladies a group of 5 is to be formed.The group consists of a president, a vice president and three secretaries. How many no. of arrangements are possible?

1)the ans is 45C7
2) there are 8C5 ways of selecting them and 5!/3! ways of arranging them...hence total is 8C5*(5*4)

[gautam_kanu]

Originally Posted by MaskedMenace

Marbles are considered identical unless mentioned.

a+b+c+d+e+f = 10

C(10-1,6-1) = 9c5

General approach is

coefficient of x^10 in (x+x^2 + x^3 + ..+x^10)^6

(1-6x^11)*(1-x)^-6

coef of x^4 in (1-x)^-6 => 4+6-1c6-1 = 9c5

Hello Masked Menace, can you throw some light on the General Approach...

[gunner4life]

Originally Posted by shanks4mba

1)the ans is 45C7
2) there are 8C5 ways of selecting them and 5!/3! ways of arranging them...hence total is 8C5*(5*4)

Approach to the first question?

[shanks4mba]

Originally Posted by gunner4life

Approach to the first question?

1) as there are 9 stops hence the total no of different types of tickets will be 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45. amongst which you gotta select any 7. hence total ways will be 45C7

[gautam_kanu]

Originally Posted by gunner4life

1. A train running between A and B has 9 stops in between. 7 passengers enter the train somewhere in between A and B There are no restrictions on no. of tickets that can be purchased on a station. In how many ways can the tickets be purchased?

2. From 4 gents, and 4 ladies a group of 5 is to be formed.The group consists of a president, a vice president and three secretaries. How many no. of arrangements are possible?

For the first question, First of all find the total number of tickets possible.
Since the passengers are boarding after A there are 10 stations to choose from(Including B)
So possibilities of tickets = 10C2 = 45

Now here comes the doubt. If more than one passanger than take the same type of ticket then the answer should be = 45^7
But if there can be no repetation of tickets answer will be 45C7(selection of 7 tickets from a lot of 45)
Guess the source has assumed that there can be no repetation of tickets, but i feel like according to the question this is not clear, so 45 ^7 should be the answer....

[Pinto]

I did'nt get this one.....
Five balls of different colours are to be placed in three boxes of different sizes.Each box can hold all five balls, The no of ways in which we can place the balls in the boxes so that no box remains empty??

[subash_r]

Originally Posted by Pinto

I did'nt get this one.....
Five balls of different colours are to be placed in three boxes of different sizes.Each box can hold all five balls, The no of ways in which we can place the balls in the boxes so that no box remains empty??

Whats the answer ...

is it 1080 ??

[shanks4mba]

Originally Posted by Pinto

I did'nt get this one.....
Five balls of different colours are to be placed in three boxes of different sizes.Each box can hold all five balls, The no of ways in which we can place the balls in the boxes so that no box remains empty??

is it 243 ways?

[KAC8L591]

Originally Posted by shanks4mba

is it 243 ways?

its should be less than 243..........

[KAC8L591]

Originally Posted by subash_r

Whats the answer ...

is it 1080 ??

is the ans 54