Question

question on Combinatorics
if : n+1 C n-2 =28
find: (n-7)!

Answer 1

what is r on r-2 ??

Answer 2

r is subset size

Answer 3

hello

Answer 4

hello

Answer 5

I need the original problem or a snapshot, please

Answer 6

yea

Answer 7

I am sorry this is the original problem . i haven't a snapshot . i am sorry Loser66

Answer 8

for the lols

Answer 9

I'm sure there's an analytic way to approach this, but an easy way to do it is to examine Pascal's triangle and seeing for which values \(\dbinom nk=28\). This occurs for \(n=8\) and \(k=2\) or \(k=6\).

For reference, see the arrays here: http://oeis.org/A007318/table

Answer 10

HolsterEmission. leave this problem .
you can solve this problem :
n+1 C n-2 =28
find: (n-7)!

Answer 11

That one's a bit easier, but are you sure it's correct? You have
\[\binom{n+1}{n-2}=\frac{(n+1)!}{(n-2)!((n+1)-(n-2))!}=\frac{(n+1)(n)(n-1)}{6}=28\]but this doesn't have integer solutions for \(n\)...

Answer 12

HolsterEmission. I ARRIVED TO THOSE STEPS BUT THEN I CAN'T CONTINUE LIKE YOU . I AM VERRY SORRY TO waste your time,HolsterEmission.

Answer 13

If you have any questions I'll be happy to explain in more detail.

Answer 14

Thank you, HolsterEmission. if i have any question .i will ask you about it .

Answer 15

To me, there is no such n satisfies the condition because there is no integer n such that
\[\left(\begin{matrix}n+1\\n-2\end{matrix}\right)=28\]

Answer 16

As what @HolsterEmission did before, the LHS = \(\dfrac{(n-1)n(n+1)}{6}=28\)
That gives us \((n-1)n(n+1) =168\)
but (n-1), n, (n+1) are 3 consecutive numbers, and 128 = 17*16 only.
If we solve the cubic, we get n = 5.5..., it is not an integer.