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# A table which has 7seats, 4 being on one side facing the

Author Message
Senior Manager
Joined: 06 Dec 2003
Posts: 366

Kudos [?]: 14 [0], given: 0

Location: India
A table which has 7seats, 4 being on one side facing the [#permalink]

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07 Feb 2004, 23:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A table which has 7seats, 4 being on one side facing the window and 3 being on the opposite side. In how many ways can 7 people be seated at the table where
1. if 2 people say, X and Y, must seat on the same side
2. X and Y must seat on the opposite side
3. if 3 people, X Y and Z must sit on the side facing the window.

Answers: (1.) 2160 (2) 2880 (3) 576

Any explanation ?

Dharmin

Kudos [?]: 14 [0], given: 0

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4285

Kudos [?]: 527 [1], given: 0

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08 Feb 2004, 00:41
1
KUDOS
1)
If X and Y sit on the side with 3 chairs, then possible outcomes are:
3C2 ( ways of picking 2 seats out of the 3 chairs ) * 2 ( when X and Y are interchanged ) * 5! ( total possibility of seating remaining 5 people ) = 3 * 2 * 5! = 6 * 5!
If X and Y sit on the side with 4 chairs, then possible outcomes are:
4C2 ( ways of picking 2 seats out of the 4 chairs ) * 2 ( when X and Y are interchanged ) * 5! ( total possibility of seating remaining 5 people ) = 6 * 2 * 5! = 12 * 5!
Total outcome = 6 * 5! + 12 * 5! = 18 * 5! = 2160

2)
If X and Y sit on different sides then:
3C1 ( ways of picking 1 seat out of 3 chairs ) * 4C1 ( ways of picking 1 seat out of 4 chairs ) * 2 ( when X and Y are interchanged ) * 5! ( total possibility of seating remaining 5 people ) = 3 * 4 * 2 * 5! = 24 * 5! = 2880

3) If X, Y, Z sit on the side facing the window ( side with 4 chairs ):
4C3 ( ways of picking 3 seat out of 4 chairs ) * 3! ( total possibility of seating X, Y and Z ) * 4! ( total possibility of seating remaining 4 persons ) = 4 * 3! * 4! = 576
_________________

Best Regards,

Paul

Kudos [?]: 527 [1], given: 0

08 Feb 2004, 00:41
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