4 Responses to “In how many ways can the executive director select 9 people if he wants to employ at least one mathematician?”

1. SILAS on November 2nd, 2009 5:48 am

   RODERICK

   There may be a way to make this count directly, but I find it easier to do it a little indirectly.

   How many groups of 9 people can be hired without the restriction of requiring 1 mathematician?
   14 C 9

   So we change the question for a moment. How many of these groups _fail_ the requirement?
   Well, this is true if all 9 are hired from the pool of 10 accountants, so there are 10 C 9 ways to fail the requirement.

   To get the answer, we simply subtract the number of ways we can fail from the total number we found up there. The number we want then is:
   14C9 - 10C9 = 2002 - 10 = 1992 ways

2. BARRY on November 3rd, 2009 10:33 pm

   VALENTIN

   total ways to choose 9 out of 14 = 14C9
   total ways to choose 9 without mathematician = 10C9
   total ways to choose 9 with >=1 mathematician =
   14C9 - 10C9

3. LEO on November 5th, 2009 6:16 pm

   LUCAS

   AT LEAST ONE MATHEMATICIAN, therefore
   one ore more

   ways = one M & remaining A + two M & remaining A+ three M & remaining A + four M & remaining A

   ways = 4 x 10C8 + 4C2 x 10C7 + 4C3×10C6+1×10C5 = 1992

   hope that helps !

4. GONZALO on November 9th, 2009 1:47 am

   SCOT

   207567360