P&C

[Neo11]

The no. of ways that 7 person can address a meeting so that three persons A,B and C from them A will speak before B and B before C is?
1680
840
120
720

[Neo11]

The no. of ways that 7 person can address a meeting so that three persons A,B and C from them A will speak before B and B before C is?
1680
840
120
720

[GMATGuruNY]

=

The no. of ways that 7 person can address a meeting so that three persons A,B and C from them A will speak before B and B before C is?
1680
840
120
720

Let the 7 people be A, B, C, D, E, F and G.

There are 7 positions to be filled.
D, E, F, and G can occupy any of the 7 positions.
Thus:
Number of options for D = 7. (Any of the 7 positions.)
Number of options for E = 6. (Any of the 6 remaining positions.)
Number of options for F = 5. (Any of the 5 remaining positions.)
Number of options for G = 4. (Any of the 4 remaining positions.)

Since A must speak before B, and B must speak before C, the 3 remaining positions must be occupied as follows: A-B-C.
Thus:
Number of options for A = 1. (A must occupy the LEFTMOST remaining position.)
Number of options for C = 1. (C must occupy the RIGHTMOST remaining position.)
Number of options for B = 1. (Only 1 position left.)

To combine all of these options, we multiply:
7*6*5*4*1*1*1 = 840.

The correct answer is B.

[vipulgoyal]

dear Mitch i think no wher mentioned a will speak just before B and B will speak just before C or not mentioned they speak consecutively


my solution

if we fix any three positions for AB&C it is clear that DEFG can be arranged in 4! ways in rest of 4 positions.

now we are considering case when ABC speaks at last three positions 1_2_3_4_5A6B7C

total ways in such case = 4! x 1 ...... Eq1

if A speaks in 4th position
B has option to speak in position 5 or 6
if B speaks on position 5, C has option to speak on position 6 or 7
and if B speaks on position 6, C has option to on position 7 only.

in same way changing the positions of A from 5 to 1st position
we may calculate in same way and that will go

(4! x 3) + (4! x 6) + (4! x 10) + (4! x 15)....Eq2

Result Eq1 + Eq2

{(4! x 1)}+{(4! x 3) + (4! x 6) + (4! x 10) + (4! x 15)}

= 840

[Brent@GMATPrepNow]

The no. of ways that 7 person can address a meeting so that three persons A,B and C from them A will speak before B and B before C is?
1680
840
120
720

Here's another approach.

First determine the number of arrangements if we ignore the restriction that A speaks before B and B before C.
There are 7 speakers, so we can arrange them in 7! ways.

Now, of course, several of these arrangements will break the restriction.
Let's examine 1 of them that breaks the restriction: AGECDBF
Here, A, B and C are speaking in the 1st, 4th and 6th positions.
In total, there are 6 ways in which A, B and C can speak in these three positions:
1) AGECDBF
2) BGECDAF
3) BGEADCF
4) CGEADBF
5) CGEBDAF
6) AGEBDCF

There are six arrangements because we can arrange these 3 people in 3! ways.

IMPORTANT: Of the 6 different arrangements where only A, B and C are moved, only one is such that A speaks before B and B before C. In other words, only 1 in 6 arrangements will conform to the restriction.

So, to find the total number of arrangements that conform to the restriction, take 7! and divide by 6

7!/6 = [spoiler]840 = B[/spoiler]

Cheers,
Brent

[GMATGuruNY]

dear Mitch i think no wher mentioned a will speak just before B and B will speak just before C or not mentioned they speak consecutively

My solution above does not restrict A, B, and C to consecutive slots.
After counting the number of options for D, E, F and G, I stated the following:

Since A must speak before B, and B must speak before C, the 3 remaining positions must be occupied as follows: A-B-C.
Thus:
Number of options for A = 1. (A must occupy the LEFTMOST remaining position.)
Number of options for C = 1. (C must occupy the RIGHTMOST remaining position.)
Number of options for B = 1. (Only 1 position left.)

To illustrate:

XDEXFXG
Here:
A must occupy slot 1, the LEFTMOST position available.
C must occupy slot 6, the RIGHTMOST position available.
B must occupy slot 4, the only slot left.

DXXEFGX
Here:
A must occupy slot 2, the LEFTMOST position available.
C must occupy slot 7, the RIGHTMOST position available.
B must occupy slot 3, the only slot left.

In each case:
A must occupy the LEFTMOST slot available.
C must occupy the RIGHTMOST slot available.
B must occupy the only slot left.
Thus, after D, E, F and G are placed, there is only 1 option for A (the leftmost slot), 1 option for C (the rightmost slot), and 1 option for B (the only slot left).
These options do NOT restrict A, B and C to consecutive slots.

[vipulgoyal]

got it,very well explained later on