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You'll solve this equation involving combinations using the factorial formula for combinations of n elements taken k at a time:
C(n,k) = n!/k!(n - k)!
Let's evaluate C(n+1 , 3) = (n+1)!/3!(n+1-3)! = (n+1)!/3!(n - 2)!
Let's evaluate C(n , 2) = n!/2!(n-2)!
Now, we'll equate the equivalent expressions:
(n+1)!/3!(n - 2)! = n!/2!(n-2)!
We'll simplify both sides by (n-2)!
(n+1)!/3! = n!/2!
We can write (n+1)! = n!*(n+1)
We can write 3! = 2!*3
n!*(n+1)/2!*3 = n!/2!
We'll simplify both sides by n!/2!:
(n+1)/3 = 1
We'll cross multiply:
n + 1 = 3
n = 3 - 1
n = 2
Since the value of "n" has to be a natural number, therefore a positive integer, we'll accept n=2 as solution of the given equation.
the answer is NOT 5!
C(a,b) may also be written as aCb, implying a!/b!(a-b)!
Hence, (n+1)C3 = (n+1)!/3!(n+1-3)!
But (n+1)! = (n+1)n!
And (n+1-3)! = (n-2)!
Also, nC2 = n!/2!(n-2)!
Thus, (n+1)n!/3!(n-2)! = n!/2!(n-2)!
At this step, multiply both sides of the equation with (n-2)!/n!
This leaves us with the equation:
(n+1)/3! = 1/2!
Multiply both sides by 3!, giving us:
n+1 = 3!/2!
Hence, n+1 = 3
And n = 3-1;
n = 2
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