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Hugo and Viviana work in an office with eight other co-workers. Out of these 10 workers, their boss needs to choose a group of four to work together on a project.

a. How many different working groups of four can the boss choose?

b. Suppose Hugo and Viviana absolutely refuse, under any circumstances, to work together. Under this restriction, how many different working groups of four can be formed ?

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Is this is correct

a. P(n,r)=n!/(n-r)!

P(10,2)=10!/(10-2)!

a. How many different working groups of four can the boss choose?

b. Suppose Hugo and Viviana absolutely refuse, under any circumstances, to work together. Under this restriction, how many different working groups of four can be formed ?

--------------------------

Is this is correct

a. P(n,r)=n!/(n-r)!

P(10,2)=10!/(10-2)!

permutation distinguishes different orderings

combination does not

Scott, Tom, Sally, Sue is the same group as Sally, Tom, Sue, Scott

right?

if so, that's a combination

Does the boss distinguish those groups?

if the boss picks Scott, Tom, Sally, Sue for a group.

is that different than if he picks Sally, Tom, Sue, Scott?

it's the same 4 people. I simply listed them in different order.

Does that matter?

if it does, then you want permutations

if it does not, then you want combinations

C(n,r) = n!/(r!(n-r)!) --- sorry I had a typo and misplaced a factorial (!) in my first post

permutation:

P(n,r) = n!/(n-r)!

a. How many different working groups of four can the boss choose?

C(n,r) = n!/(r!(n-r)!) = 210

part b is already posted

how is http:#37033589 an answer?

first, it's just a question

second, it's a repeat question of http:#37033507

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then for part B

C(8,4) + C(8,3) + C(8,3) = 182

C(8,4) = all combinations of 4 employees that don't include either Hugo or Viviana

C(8,3) = all combinations of 3 employees that could partner with Hugo (except Viviana)

C(8,3) = all combinations of 3 employees that could partner with Viviana (except Hugo)