Question

im trying to solve P(n,3) = 2(n-1 P 3) using permutations

Answer 1

and where are u stuck ?

Answer 2

how to solve each side to get an answer

Answer 3

you know the definition of \(\large _nP^r\) ?

Answer 4

sorry,
\(\large ^nP_r\)

Answer 5

i know its permutations

Answer 6

so,
\(\huge \dfrac{n!}{(n-3)!}=2 \times \dfrac{(n-1)!}{(n-4)!}\)
did u get till here ?

Answer 7

yes, i don't know how to show to solve each side from there

Answer 8

use the fact that \(\huge n!=n(n-1)!\)
and
\(\huge n!=n(n-1)!=n(n-1)(n-2)!\)
and so on...

Answer 9

how do i solve the right side of the right side f the equation when its multiplied by 2?

Answer 10

\(\huge \dfrac{n(n-1)!}{(n-3)(n-4)!}=2 \times \dfrac{(n-1)!}{(n-4)!}\)
did you get what i did ?

Answer 11

numerator of left is obvious,
for denominator, \(\large (n-3)! = (n-3)(n-3-1)! = (n-3)(n-4)!\)
got this ?

Answer 12

i dont understand why the denominator is like that

Answer 13

\(\large a!=a(a-1)!\)
this is standard, right ?
now just replace a by n-3

Answer 14

ok

Answer 15

now what gets cancelled ?

Answer 16

i dont know

Answer 17

now you got this right ?
\(\huge \dfrac{n(n-1)!}{(n-3)(n-4)!}=2 \times \dfrac{(n-1)!}{(n-4)!}\)
how can you simplify this ? notice that some terms are getting cancelled!

Answer 18

does the (n-1)! on the tops cancel out and the (n-4)! on the bottoms cancell out?

Answer 19

YES!
and what remains ??

Answer 20

\[\frac{ n }{ (n-3) } = 2\]

Answer 21

yes, isn't this easy to solve ?
(NOTE : since we are cancelling (n-1)! , which can be 0, we need to account for n-1 =0 and so, n=1 is a solution)

Answer 22

and since we cancelled, (n-4) ! also, solve (n-4)! = 0, n=.... ?, ...?

Answer 23

are u getting all this ?

Answer 24

so becasue it equaled the fraction, then it goes down to 2(n-3)=n?

Answer 25

and then it goes to 2n-6=n ?

Answer 26

yes, continue, n=... ?

Answer 27

2n=n+6 ?

Answer 28

is that right?

Answer 29

or is it n =6

Answer 30

thanks for your help!

Answer 31

n=6 is correct! :)
welcome ^_^
and
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