Problem Solving

Rahul@gurome wrote:Fix one person in one place, so the permutations for the remaining 4 will be 4! = 4*3*2*1 = 24

[spoiler]The correct answer is (C).[/spoiler]

Sorry, I am unable to understand your explanation.

How is this question different from normal seating arrangement problem.

In normal seating arrangement problem answer would be 5!

lokesh r wrote:

Rahul@gurome wrote:Fix one person in one place, so the permutations for the remaining 4 will be 4! = 4*3*2*1 = 24

[spoiler]The correct answer is (C).[/spoiler]

Sorry, I am unable to understand your explanation.

How is this question different from normal seating arrangement problem.

In normal seating arrangement problem answer would be 5!

Let's say that the 5 people are ABCDE.

If we were to count the ways to arrange ABCDE in a line, the following would qualify as different arrangements:

ABCDE
BCDEA
CDEAB
DEABD
EABCD

But when put around a table, all of the above would qualify as only one arrangement, because the clockwise order would always be the same: A-B-C-D-E. In all of the above, B would be directly to the right of A; C would be directly to the right of B; D would be directly to the right of C; and E would be directly to the right of D.

Thus, the number of ways to arrange N people around a circular table is smaller than the number of ways to arrange them in a line:

Number of ways to arrange N people around a circular table = (N-1)!.

So given 5 people, there are (5-1) = 4! = 24 ways to arrange them around a table.

GMATGuruNY wrote:

lokesh r wrote:

Rahul@gurome wrote:Fix one person in one place, so the permutations for the remaining 4 will be 4! = 4*3*2*1 = 24

[spoiler]The correct answer is (C).[/spoiler]

Sorry, I am unable to understand your explanation.

How is this question different from normal seating arrangement problem.

In normal seating arrangement problem answer would be 5!

Let's say that the 5 people are ABCDE.

If we were to count the ways to arrange ABCDE in a line, the following would qualify as different arrangements:

ABCDE
BCDEA
CDEAB
DEABD
EABCD

But when put around a table, all of the above would qualify as only one arrangement, because the clockwise order would always be the same: A-B-C-D-E. In all of the above, B would be directly to the right of A; C would be directly to the right of B; D would be directly to the right of C; and E would be directly to the right of D.

Thus, the number of ways to arrange N people around a circular table is smaller than the number of ways to arrange them in a line:

Number of ways to arrange N people around a circular table = (N-1)!.

So given 5 people, there are (5-1) = 4! = 24 ways to arrange them around a table.

Thanks for the explanation..all these days i only knew formula..