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Please help solve this Combinatorics problem

Expert replies
by dddanny2006 » Thu Dec 05, 2013 3:33 pm
Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many ways different arrangements can the 6 people sit?


Please solve this via the slot method.Thanks
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Source: — Problem Solving |

by [email protected] » Thu Dec 05, 2013 9:06 pm
Hi dddanny2006,

Since the GMAT tends to "reward" flexible thinkers, it's important to be comfortable thinking logically and in a number of different ways. I'm going to show you a way to think of this question that's more about logic and less about math.

Since there are 6 people sitting in 6 adjacent seats, using permutation logic/math makes sense.

IF there were NO restrictions, then the total number of options would be: 6x5x4x3x2x1 = 720 arrangements.

BUT there is a restriction: Marcia and Jan CAN'T sit next to one other. That means we'll have to subject some of the options away from the 720.

Imagine if we put Marcia in seat 1 and Jan in seat 2. Then we'd have....

MJ4x3x2x1 = 24 ways with Marcia 1st and Jan 2nd.

If we flipped those two around, we'd have...

JM4x3x2x1 = 24 ways with Jan 1st and Marcia 2nd.

24 + 24 = 48 ways that DON'T WORK if we put Marcia and Jan in the first 2 spots.

We can use that SAME pattern throughout the row:

2nd and 3rd = 48 options that DON'T WORK
3rd and 4th = 48 options that DON'T WORK
4th and 5th = 48 options that DON'T WORK
5th and 6th = 48 options that DON'T WORK

In total, there are 48(5) = 240 ways that DON'T WORK. Subtract those ways from the total possible.

[spoiler]720 - 240 = 480 arrangements.[/spoiler]

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by theCodeToGMAT » Thu Dec 05, 2013 10:28 pm
6! = Total ways to arrange 6 people
5!2! = Considering Marcia and Jan in a single group and other 4 in each seperate group

Total ways = 6! - 5!2! = 5!(6 - 2) = 480
R A H U L
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by GMATGuruNY » Fri Dec 06, 2013 3:52 am
dddanny2006 wrote:Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many ways different arrangements can the 6 people sit?
Good arrangements = total arrangements - bad arrangements.

Total arrangements:
Number of ways to arrange the 6 elements G, M, P, J, B and C = 6! = 720.

Bad arrangements:
In a BAD arrangement, M and J are in adjacent positions.
Put M and J together in a BLOCK, so that MJ serves as SINGLE ELEMENT in the arrangement.
Number of ways to arrange the 5 elements MJ, G, P, B and C= 5! = 120.
in each of these arrangements, MJ can be reversed to JM, DOUBLING the total number of options.
Thus, we multiply by 2:
2*120 = 240.

Good arrangements:
Total-bad = 720-240 = 480.
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by Mathsbuddy » Fri Dec 06, 2013 7:52 am
000011
000110
001100
011000
110000

The above presents 5 illegal positions * 2 variations (J & M could swap seats) * 4! arrangements of the other 4 seats

6 * 2 * 4! = 240 different illegal arrangements.

Out of 6! = 720 possibilities, this leaves

720 - 240 = 480 arrangements.
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