Problem9 Combinations, Subsets, and Partitions) Suppose you have a committee of

Question

Problem9 Combinations, Subsets, and Partitions) Suppose you have a committee of 10 people. (a) How many ways are there to choose a group of 4 people from these 10 if two particular people (say, John and Dave) can not be in the group together? (b) How many ways are there to choose a team of 5 people from these 10 with one particular person being designated Captain and another particular person being designated Co-Captain? (c) How many ways are there to separate these 10 people into two groups, if no group can have less than 2 people? (d) How many ways are there to separate these 10 people into two teams of 6 and 4 people? (e) How many ways are there to separate these 10 people into four groups of 2, 2, 3, and 3 each? Hint: For (c) - (e), think about whether or not you are... (wait for it.. over-counting.

Explanation / Answer

a)
total number of ways without any constraint = 10C4 = 210
now
number of ways to select 4 people out of 10 when two particular people are together
= 8C2   {note that 2 are already selected }
= 28

hence
required number = 210 - 28 = 182

b)
so two person are already determined
hence we have to select 5-2 = 3 people out of 10-2 = 8
hence
8C3 = 8*7*6/6 = 56

c)
x1 + x2 = 10
x1 >= 2 and x2 >= 2
x1 = y1+ 2 , x2 = 2 +y2
hence
x1 + x2 = 10   reduces to
y1 + y2 = 6 where yi >= 0
number of non-negative integral solution
y1 + y2+ ..yr = n is (n+ r -1)C(r-1)

hence number of ways = (6 + 2-1)C(2-1) = 7C1 = 7

d)
we have to select 6 people out of 10
= 10C6 ways
= 10C4 = 210 ways
{note that the other 4 left will form separate team automatically}

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