SOLUTION: nCr:nC(r-1) = 2:3 , nC(r-2):nC(r-1) = 4:3 find n and r Could someone solve it for me

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Question 1044457: nCr:nC(r-1) = 2:3 , nC(r-2):nC(r-1) = 4:3 find n and r
Could someone solve it for me
Found 2 solutions by Boreal, Edwin McCravy:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
nCr=n!/r!(n-r)!
nC(r-1)=n!/(r-1)!(n-r+1)!, watch the sign there.
nC(r-2)=n!/(r-2)(n-r+2)!
-------------------Divide nCr by nC(r-1)
n!/r!(n-r)!/n!/(r-1)!(n-r+1)! The n! cancels and you invert the denominator
(n-r+1)!(r-1)!/r!(n-r)!
(r-1)!/r!=1/r
(n-r+1)!/(n-r)!=n-r+1
The quotient for this is (n-r+1)/r=2/3
cross-multiply and you get 3n-3r+3=2r
3n+3=5r
------------------------
now do nC(r-2)/nC(r-1)
n!/(r-2)!(n-r+2)!/n!/(n-r+1)!
do the same thing with canceling the n! and inverting the denominator
get (r-1)!(n-r+1)!/(r-2)!(n-r+2)!
this is (r-1)/(n-r+2) =4/3
cross-multiply
3r-3=4n-4r+8
7r=4n+11
---------------------
rewrite as
7r-4n=11
5r-3n=3
multiply the top by 3 and the bottom by (-4)
21r-12n=33
-20r+12n=-12
r=21
substitute and n=34
34C21, 34C20, 34 C 19
The first is 927983760, the second is 1391975640, and they are in a 2:3 ratio. Really!
The third is 1855967520
The last divided by the second is 1.3333 repeat, which is 4:3
n is 34
r is 21
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
nCr:nC(r-1) = 2:3 , nC(r-2):nC(r-1) = 4:3 find n and r

Simplify the first equation:







Divide both sides by n!





In the right side, write r! as r(r-1)!



Divide both sides by (r-1)!



Write (n-r+1)! as (n-r+1)(n-r)!



Divide both sides by (n-r)!







Looks better turned around:



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

nC(r-2):nC(r-1) = 4:3 

Simplify the second equation:









Divide both sides by n!





In the left side, write (r-1)! as (r-1)(r-2)!



Divide both sides by (r-2)!



On the right side, write (n-r+2)! as (n-r+2)(n-r+1)!



Divide both sides by (n-r+1)!








Now we solve the system of two equations for r and n



I'll let you solve that by elimination (addition). 

n = 34, r = 21

Checking in the first given proportion:

nCr:nC(r-1) = 2:3 , nCr-2:nCr-1 = 4:3

34C21:34C20 = 927983760:1391975640

Both those huge integers are divisible by the
huge integer 463991880

927983760/463991880 = 2 and 1391975640/463991880 = 3

So that checks.

Checking n = 34, r = 21 in the second given proportion:

nC(r-2):nC(r-1) = 4:3

34C19:34C20 = 1855967520:1391975640

Both those huge integers are divisible by the
huge integer 463991880

1855967520/463991880 = 4 and from above 1391975640/463991880 = 3

That also checks.

Edwin
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