Asked by brina
Say you have 12 books, but room to display only 7. I will open this to everyone - how many differnt arrangements are there if the order of the books matters (permutation) and how many if order is of no consequence (combination)?
Answers
Answered by
Anonymous
lajnklh
Answered by
Reiny
If the order does matter than the number of permutations is 12*11*10*9*8*7*6= P(12,7) = 3991680
if the order does not matter (combinations)
it would be 12!/(7!5!) = C(12,7) = 792
both P(n,r) and C(n,r) can be found on a standard scientific calculator
usually they are labeled <sub>n</sub>P<sub>r</sub> and <sub>n</sub>C<sub>r</sub>
to do C(12,7)
enter
12
2nd F
<sub>n</sub>C<sub>r</sub>
7
=
you should get 792
if the order does not matter (combinations)
it would be 12!/(7!5!) = C(12,7) = 792
both P(n,r) and C(n,r) can be found on a standard scientific calculator
usually they are labeled <sub>n</sub>P<sub>r</sub> and <sub>n</sub>C<sub>r</sub>
to do C(12,7)
enter
12
2nd F
<sub>n</sub>C<sub>r</sub>
7
=
you should get 792
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