Solving word problems

Question

Ahmad is on a Facebook group that has 7 members. How many different sub groups of 3 people could be formed that have Ahmad as a members?

Answer 1

You want to make a group of 3 out of a group of 7.

 

Think about filling 3 "slots" with people's name from the group of 7 members.   You have 7 choices for the first slot, once you have chosen that name, you have 6 names left to choose from for the second slot, and then 5 left for the third slot.  So you have 7x6x5 choices to fill those 3 slots.  But whenever you are dealing with a group of people, say if you and me and Ahmad were the 3 names chosen - it would be the same group of 3 people if they chose you first, then Ahmad, then me.  Or if they chose me first, then you then Ahmad.  So by calculating 7x6x5, we have overcounted the distinct number of groups.

 

So we have to divide this number by the number of ways you can make a group of 3 out of 3 things.  Think of three slots.  You have 3 choice for the first slot, then after filling the first, you have 2 choices for the second slot, then 1 left for the third slot. So you have 3x2x1 ways in all to arrange the 3 things once you have chosen that group of 3.  

 

So we divide the total number of groups of 3 that can be formed from these 7 people - 7x6x5 by the number of ways you can arrange any 3 people - 3x2x1.

 

The answer is (7x6x5)/(3x2x1)

 

There's a button on your calculator that you can use - but you have to be VERY careful that you know the difference between PERMUTATIONS and COMBINATIONS - but you can google the nPr and nCr buttons on your calculator to find out more.  Because of the confusion that arises, I always like to make sure my SAT students know the reasoning behind the calculations, - and then they can do these problems even without the calculator.

Answer 2

The answer felt a little high to me, so I double-checked my instincts empirically and found only 15 sets with Ahmad included. This is consistent with the following:

While combinations are indeed the key, requiring him to appear in every grouping means you can only choose from among 6 people in groups of 2 (before Ahmad makes three). This means we perform this calculation:

₆C₂ =   6!          = 6!        = 6*5*4*3*2*1   

          (6-2)! 2!       4! * 2!    4*3*2*1*2*1

Note that the "4*3*2*1" cancels on top and bottom, leaving:

6 * 5  = 3*5 = 15

 2 * 1