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sehosayho
A conference is to have eight presentations over the course of one day, consisting of three long presentations, and five short presentations. If the conference organizer doesn't want consecutive long presentations, and the conference is to start with a short presentation, how many schedules of presentations are possible?

A. 120
B. 720
C. 2880
D. 5760
E. 11520

the book says the answer is 2880 which I dont agree with.

I have the possible ways of 5short and 3long = 720.
then the possible arrangement = 7
so I got 5040.

To find the arrangement, i had to draw squares and fill then with the symbols of Ss and Ls.

Anyone knows better ways or equation for this?

S*S*S*S*S*

5 short presentations can be arranged in 5!=120 ways.

As for 3 long presentations: each can be placed instead of any of the 5 star, so in \(C^3_5=10\) ways, and they also can be arranged in 3!=6 ways.

Total = 120*10*6 = 7,200.

No correct answer among the options.
Bunuel, can you give a solution with the opposite approach, total minus always together. I am not able to get the approach correctly.

Thanks
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Bunuel
sehosayho
A conference is to have eight presentations over the course of one day, consisting of three long presentations, and five short presentations. If the conference organizer doesn't want consecutive long presentations, and the conference is to start with a short presentation, how many schedules of presentations are possible?

A. 120
B. 720
C. 2880
D. 5760
E. 11520

the book says the answer is 2880 which I dont agree with.

I have the possible ways of 5short and 3long = 720.
then the possible arrangement = 7
so I got 5040.

To find the arrangement, i had to draw squares and fill then with the symbols of Ss and Ls.

Anyone knows better ways or equation for this?

S*S*S*S*S*

5 short presentations can be arranged in 5!=120 ways.

As for 3 long presentations: each can be placed instead of any of the 5 star, so in \(C^3_5=10\) ways, and they also can be arranged in 3!=6 ways.

Total = 120*10*6 = 7,200.

No correct answer among the options.

Baffled with the pattern of Q. I find out following combinations;

SLSLSLSS
SSLSLSLS
SSSLSLSL

Can't find the way to solve the problem.

:shock:
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rango
Bunuel
sehosayho
A conference is to have eight presentations over the course of one day, consisting of three long presentations, and five short presentations. If the conference organizer doesn't want consecutive long presentations, and the conference is to start with a short presentation, how many schedules of presentations are possible?

A. 120
B. 720
C. 2880
D. 5760
E. 11520

the book says the answer is 2880 which I dont agree with.

I have the possible ways of 5short and 3long = 720.
then the possible arrangement = 7
so I got 5040.

To find the arrangement, i had to draw squares and fill then with the symbols of Ss and Ls.

Anyone knows better ways or equation for this?

S*S*S*S*S*

5 short presentations can be arranged in 5!=120 ways.

As for 3 long presentations: each can be placed instead of any of the 5 star, so in \(C^3_5=10\) ways, and they also can be arranged in 3!=6 ways.

Total = 120*10*6 = 7,200.

No correct answer among the options.

Baffled with the pattern of Q. I find out following combinations;

SLSLSLSS
SSLSLSLS
SSSLSLSL

Can't find the way to solve the problem.

:shock:

Notice that all S's and all L's there are different and re-read my post.

Hope it helps.
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sehosayho
A conference is to have eight presentations over the course of one day, consisting of three long presentations, and five short presentations. If the conference organizer doesn't want consecutive long presentations, and the conference is to start with a short presentation, how many schedules of presentations are possible?

A. 120
B. 720
C. 2880
D. 5760
E. 11520

the book says the answer is 2880 which I dont agree with.

I have the possible ways of 5short and 3long = 720.
then the possible arrangement = 7
so I got 5040.

To find the arrangement, i had to draw squares and fill then with the symbols of Ss and Ls.

Anyone knows better ways or equation for this?


You mean that you first find out the no. of ways S selection, then ways of L selection, then the ways of arrangements of S and L and atlas multiply all the ways to get the final answer.

Thanks Bunuel

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