+0

# help asap, thanks a lot

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Let \(n\) and \(k\) be positive integers such that\(\binom{n}{k}:\binom{n}{k + 1} = 4:11.\)
Find the smallest possible value of \(n\).

Mar 27, 2020

#1
+111330
+1

We can simplify this as

n!                     ( n - [k + 1])! (k + 1)!

_______   *       ________________       rearrange  as

(n - k!) k!                   n!

(n - k - 1) !          (k +1)!                    k  + 1

________  *      ______    =            _____

(n - k)!              k!                          n  -  k

And we know that

k+ 1           4

____  =   ____        cross-multiply

n - k         11

11(k + 1)  =  4 ( n - k)

11k + 11 = 4n - 4k

15 k  =  4n  - 11

15k + 11

_______   =  n

4

n will be  smallest    when  k =  3

So n  =  14

Proof

C ( 14, 3)          364            4

________  =    ____   =    ___

C ( 14, 4)         1001          11

Mar 27, 2020
#2
+1

thanks so much, really appreciate it :)

Mar 27, 2020
#3
+24995
+1

Let \(n\) and \(k\) be positive integers such that \(\dbinom{n}{k}:\dbinom{n}{k + 1} = 4:11\). Find the smallest possible value of \(n\).

Mar 27, 2020