In how many ways the numbers 1, 2, 3, 4, 5, and 6 can be arranged arou

Question

In how many ways the numbers 1, 2, 3, 4, 5, and 6 can be arranged around a circle so that the odd and even numbers are altering? (2 seating arrangements are considered different only when the positions of the people are different relative to each other.)

A. 12
B. 24
C. 36
D. 108
E. 120

A. 12
B. 24
C. 36
D. 108
E. 120

GMATinsight · Answer

Let's mark 6 positions around the circle as a, b, c, d, e, f

Let's fix a, c and e with Odd numbers  
 Let's fix b, d and f with Even numbers

We need to fix the position of one of the numbers so that the orders are not repeated

Let's assume that we fix number 1 at 'a'

Remaining two odd numbers can be arranged at remaining two places fixed for them in 2! ways

Three Even numbers can be arranged at their places fixed for them in 3! ways

Total Arrangements =  2! * 3!  = 12

Answer: Option A

deepfuknchange · Answer

No solution here, but I started manually finding how many distinct combinations starting with 1, ending with a 2/4/6.
After 2 minutes I guessed between 12 or 24, figuring I could multiply by 3 to capture combinations starting with 3 ending 2/4/6, and starting with 5 ending 2/4/6.

Without knowing how to set up the formula(s) hopefully the logic gave me a 50-50 chance to be correct.

suntprovident · Answer

GMATinsight  why don't I multiply by 2 to consider cases when say even nos are fixed. Currently say Odd nos are occupying the 1st, 3rd and 5th chair ..

KarishmaB · Answer

Visualize 6 identical chairs around a table. Pick one of the odd numbers, say 3, and place in anywhere in one way only since all chairs are identical. 
Now all chairs have become distinct relative to the number 3. There are 2 distinct seats for the 2 odd numbers (the other odd numbers cannot be next to 3 and there must be a chair between them so they can occupy only the 2 alternately placed chairs.) 
The 2 odd numbers can be placed in 2! ways.

For the 3 even numbers, there are 3 distinct chairs so they can be placed in 3! ways.

Total = 2! * 3! = 12

Answer (A)

Bunuel · Answer

When you arrange n objects in a circle, the number of arrangements is (n−1)!, because rotating everyone by the same amount does not create a new arrangement. The relative order stays the same.

So for the 3 odd numbers, the number of arrangements is 2!.

After the odd numbers are placed, the circle is effectively fixed. The 3 spots for the even numbers are the 3 specific gaps between consecutive odd numbers.

Now the even numbers are not “in their own circle.” They are being placed into 3 fixed positions relative to the odd numbers. If you shift all the even numbers by one gap, each even number ends up between different odd numbers, so it is a different overall arrangement.

Therefore, the number of ways to place the 3 even numbers is 3!, not 2!.

Thus, the total number of arrangements is 2! * 3! = 12.

Hope it's clear.

In how many ways the numbers 1, 2, 3, 4, 5, and 6 can be arranged arou

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In how many ways the numbers 1, 2, 3, 4, 5, and 6 can be arranged arou

Prep Toolkit

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