mustish1
asked on
Permutation
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)!
ASKER
How to know that its a combination or permutation ?
as in your previous question
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
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
ASKER
so according to the question part a is about permutation right?
SOLUTION
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you tell me.
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
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
ASKER CERTIFIED SOLUTION
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ASKER
what are the 2 formulas for permutation and combination
combination:
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)!
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)!
ASKER
ok thanks. Can you please tell me the part a its a permutation or combination
a. How many different working groups of four can the boss choose?
a. How many different working groups of four can the boss choose?
does the order of the people in the group matter or not?
ASKER
No
if order makes a difference, its a permutation. If order makes no difference, its a combination
SOLUTION
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I don't understand the split.
how is http:#37033589 an answer?
first, it's just a question
second, it's a repeat question of http:#37033507
how is http:#37033589 an answer?
first, it's just a question
second, it's a repeat question of http:#37033507
C(n,r) = n!/(r!(n-r!))