<PRE><B>Question 1039769</B>
i believe the formula will be as follows:


there are 10 people total and you want to choose a committee of 4 out of the 10.


barry and harry are 2 people out of the 10.


if the committee is to have only barry on it, then you get to choose 4 out of 9.


if the committee is to have only harry on it, then you get to choose 4 out of 9 again.


some of those committees will have the same members on them, so you will have to subtract one occurrence of them to get the correct total.


the formula would then become 9C4 + 9C4 - 8C4.


the common members are the members of the committee that don&#39;t have barry or harry on them.


your solution should be 9C4 + 9C4 - 8C4 = 2 * 9C4 - 8C4 = 2 * 126 - 70 = 182.


this is tough to see because of all the possible committees that can be formed, but we can look at a smaller set to see if the principle works.


i did a test with 4 out of 7.


let the members be abcdefg


if a only is on the team, then the possible members become acdefg.
if b only is on the team, then the possible members becomes bcdefg.
if a and b are not on the team, then the possible members becomes cdefg.


with a only on the team, the possible combinations become 6C4.
with b only on the team, the possible combinations becomes 6C4.
with a and b not on the team, the possible combinations becomes 5C4.


the solution should be 2 * 6C4 - 5C4 = 2 * 15 - 5 = 25 possible teams.


let&#39;s see if that works.


the possible teams of 4 that can be taken from acdefg are:


---------
acde
acdf
acdg
acef
aceg
acfg
adef
adeg
adfg
aefg
--------- 
cdef
cdeg
cdfg
cefg
defg
---------


the possible teams of 4 that can be taken from bcdefg are:


---------
bcde
bcdf
bcdg
bcef
bceg
bcfg
bdef
bdeg
bdfg
befg
---------
cdef
cdeg
cdfg
cefg
defg
---------


as you can see, there are 15 teams that can be formed with only a as a possible candidate in them and there are 15 teams that can be formed with only b as a possible candidate in them.


out of each set of 15, there are 5 teams that are formed that don&#39;t have a or b in them.


those teams are common to both sets so one of them has to be removed to avoid duplicating.


15 + 15 - 5 = 25.


that&#39;s what the formula of 2 * 6C4 + 5C4 got you when the total possible combinations were 7C4 if barry and harry would just have agreed to work on the same team together.


so 7C4 became 2 * 6C4 - 5C4 because barry and harry refused to be on the same team together.


likewise 10C4 becomes 9C4 + 9C4 - 8C4 when barry and harry refuse to be on the same team together which is the formula we used above to get the total of 182 possible teams where either barry or harry is on the team, but not both.


9C4 is the number of ways you can get set of 4 out of 9.
8C4 is the number of ways you can get set of 4 out of 8.


here&#39;s a reference on combination formula.


&lt;a href = &quot;http://www.mathwords.com/c/combination_formula.htm&quot; target = &quot;_blank&quot;&gt;http://www.mathwords.com/c/combination_formula.htm&lt;/a&gt;


for example:


9C4 is equal to 9! / (4! * 5!), which is equal to (9*8*7*6*5!) / (4! * 5!), which is equal to (9*8*7*6) / (4!) because 5! cancels out, which is equal to (9*8*7*6) / (4*3*2*1), which is equal to 3024 / 24, which is equal to 126.


likewise, 8C4 becomes equal to (8*7*6*5) / (4*3*2*1) = 1680 / 24 = 70.


2 * 9C4 - 8C4 becomes equal to 2 * 126 - 70 which becomes equal to 182 as we calculated earlier, only i used the combination formula from the calculator earlier rather than doing it manually as i just did.













</PRE>