How many ways are there to separate these 10 people into two groups, if no group can have less than 2 people?

501 The number of ways of selecting n out of m is m!/(n!m!). So let's do that for selecting 2, 3, 4, 5, 6, 7, and 8 out of 10. 2!/(2!8!) = 45 3!/(3!7!) = 120 4!/(4!6!) = 210 5!/(5!5!) = 252 6!/(6!4!) = 210 7!/(7!3!) = 120 8!/(8!2!) = 45 Now add them all together. 45 + 120 + 210 + 252 + 210 + 120 + 45 = 1002 But you're not quite done. It there really any difference between having a group with persons AB and the 2nd group being CDEFGHIJ or the first group being CDEFGHIJ and the 2nd group AB? The answer is no, so you need to divide the number above by 2, giving 1002 / 2 = 501 So there are 501 to separate 10 people into 2 groups with each group having at least 2 people.

It will b. 975 because i’m guessin