1) The set of points in a plane that are equidistant from a given point form a _ _ _ _ _ _.
2 answers:
You might be interested in
Hope this helps you :)
The solution is n = 5 and r = 4
Step-by-step explanation:
Given,
nCr : (n+1)Cr : (n+2)Cr = 1:3:7
To find the value of n and r.
Formula
nCr = [ n! means = n.(n-1).(n-2)....3.2.1]
Now,
nCr : (n+1)Cr = 1:3 and (n+1)Cr : (n+2)Cr = 3:7
or, :
= 1:3 or,
:
=
or, ×
=
or,
×
=
or, =
or,
=
or, 3(n+1-r) = n+1 or, 7(n+2-r) = 3(n+2)
or, 3n+3-3r = n+1 or, 7n+14-7r = 3n+6
or, 2n-3r = -2 or, 4n-7r = -8
Now, by solving
2n-3r = -2 -----(1)
4n-7r = -8 -----(2) we will get n and r
Multiplying (1) by and then subtract with (2) we get,
2(2n-3r) - (4n-7r) = -4-(-8)
or, 4n-6r-4n+7r = 4
or, r = 4
From (1) we get,
2n = -2+3(4)
or, 2n = 10
or, n = 5
Hence,
n = 5 and r = 4