anonymous
  • anonymous
how many permutations are there of 3 things taken from a set of 7?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@amistre64
nubeer
  • nubeer
if you are using a calculator.. you can just plug in the values 7P3
amistre64
  • amistre64
the way i remember the is to start at 7 and pick 3 numbers to multiply: 7*6*5 = 7P3

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More answers

anonymous
  • anonymous
where does the 6 come from?
amistre64
  • amistre64
7 ways to pick the first element 6 ways to pick the second element 5 ways to pick the third element
amistre64
  • amistre64
if we made a tree diagram of this, it would be 7 branches with 6 branches each; 7*6 and each branch has 5 leaves: (7*6)*5 total outcomes
amistre64
  • amistre64
for any given set of elements, there are 6 ways that they can be presented: abc acb bac bca cab cba, which means that for combinations, we have 6 like sets. (7P3)/3! would be the combinatorial calculation ... but thats not really what it asked for is it :)
anonymous
  • anonymous
so confused. whats the simplest way to solve this?
amistre64
  • amistre64
start at 7 and pick 3 numbers to multiply: 7*6*5
anonymous
  • anonymous
so the answer is 210? i dont get how
anonymous
  • anonymous
how do you pick the 3 numbers?
amistre64
  • amistre64
i explained how ... which you say is confusing 6*5 = 30, 30*7 = 210
amistre64
  • amistre64
start at 7 and go down
amistre64
  • amistre64
spose you have a litter of 7 kittens; how many ways can you choose the first one to pick up? after picking up the first one, how many ways are left in which you can choose to pick up the next one? after youve picked 2 kittens, how many ways are left for you to be able to choose the last kitten?
anonymous
  • anonymous
so if it was: how many permutations are there of 2 things taken from a set of 5? the answer would be: 5*4 =20 ?
amistre64
  • amistre64
7 to choose from; then 6 to choose from, then 5 to choose from
amistre64
  • amistre64
yes 5*4 is the permutations of 2 things from 5
anonymous
  • anonymous
ohh thank youu
amistre64
  • amistre64
youre welcome
anonymous
  • anonymous
"If K objects out of a set of N elements are placed into a fixed order, there are ________ unique arrangements of those objects" ??
amistre64
  • amistre64
start with n and start picking out k elements or rather, theres some formula you can use ....\[\frac{n!}{(n-k)!}\]
anonymous
  • anonymous
huhh??
kropot72
  • kropot72
Are the N elements all different?
amistre64
  • amistre64
a set is properly defined by distinct elements ....
amistre64
  • amistre64
a set that has N elements is by definition N different elements
anonymous
  • anonymous
so the blank would be N, correct?
amistre64
  • amistre64
no
amistre64
  • amistre64
7 Pick 3 did not amount to 7
amistre64
  • amistre64
\[\frac{7*6*5*4*3*2*1}{4*3*2*1}=\frac{7!}{(7-3)!}\]
anonymous
  • anonymous
i dont get the bottom..
amistre64
  • amistre64
k elements picked from a set of n elements is usually calculated as: n!/(n-k)!
amistre64
  • amistre64
notice that in order to simplify this to 7*6*5, we have to divide out the 4*3*2*1
amistre64
  • amistre64
in general; to simplify n! to account for nPk, we need to divide out (n-k)!
anonymous
  • anonymous
ok
anonymous
  • anonymous
so....
anonymous
  • anonymous
soo lost...

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