The numbers 1,2,3,4,5,6 are arranged in a circle. How many ways can this be done if
1) at least 2 odd numbers are together
2) the numbers 5 and 6 are together but 4 is not next to 5 or 6

1. The number of possible arrangements without restrictions are 6! / 6 = 120.

Now, consider a group of odd numbers. You have only 1, 3 and 5. You can put them into groups 1, 3; 3, 5 or 1, 5, each giving 2 possible arrangements.

Now, take a group of odd numbers (1, 3), and treat them like a separate item. You therefore will have 5 items total, namely (1, 3), 2, 4, 5, 6.

The number of possible arrangements of the items is found to be 5! / 5 = 24.

Total possible arrangements = 24 x 2 x 3 = 144.

2 because you have 2 different combinations for the group of odd numbers, and 3 because you have 3 different groups containing different odd numbers.

2. Now, 5 and 6 are together: arrangements = 2
Total items = 5 (again)

(5, 6), 1, 2, 3, 4

Now, 4 must not be put next to (5, 6) you can therefore have:

1, (5, 6), 2, 3, 4
1, (5, 6), 3, 2, 4
2, (5, 6), 1, 3, 4
2, (5, 6), 3, 1, 4
3, (5, 6), 1, 2, 4
3, (5, 6), 2, 1, 4

1, 2, (5, 6), 3, 4
1, 3, (5, 6), 2, 4
2, 1, (5, 6), 3, 4
2, 3, (5, 6), 1, 4
3, 1, (5, 6), 2, 4
3, 2, (5, 6), 1, 4

That makes 12 other possible arrangements (don't forget that they are arranged in a circle.)

So, number of combinations = 2 x 12 = 24