dzyubam wrote:

See this link for more info:

7-p395941what do you all think about this approach:

8c3 are the total number of combinations

4c1 represent the choice of 1 couple od siblings

6 are the people who can be chosen to occupy the third place

8c3-4c1*6=56-24=32

moreover do you think about this other way?

Pick one person out of 8, then one out of 6 (the first person with her or his sibling excluded), then one out of 4, giving 8*6*4 permutations. These are called permutations because we count every distinct combination of 3 people 6 times (6=3!). For example, we count {Ann, Bill, and Jane} as {Ann, Bill, Jane}, {Ann, Jane, Bill}, {Jane, Ann, Bill}, {Jane, Bill, Ann}, {Bill, Ann, Jane}, {Bill, Jane, Ann}. To get the number of combinations, we divide the permutations by 6, to get (8*6*4)/6 = 32 combinations

I am not sure about its quickness.