binomials (1 Viewer)

[cormglakes]

how do i solve
nC4 + nC5 =12C5

answers : n=11

 

[cossine]

nC4 + nC5

= n! / [(4!) * (n-4)!] + n! / [ (5!) * (n-5)! ]

= 5*n! / [(5!) * (n-4)!] + n! * (n-4) / [5! * (n-4)!]

= [ 5*n! + n! * (n-4)] / [5! * (n-4)!]

= [ n! (n-4 + 5)] / [5! * (n-4)!]

= [ n! (n+1)] / [5! * (n-4)!]

= (n+1)! / [5! * (n-4)!]

Solve

(n+1)! / [5! * (n-4)!] = 12C5

=> (n+1)! = 12! * 5! * (n-4)! / (5! * 7!)

=> (n+1)! / (n-4)! = 12! */ 7!

=> (n+1) * n * (n-1) * (n -2) * (n-3) = 12 * 11 * 10 * 9 * 8

By inspection we see the expression will be equal when (n+1) = 12

Therefore n = 11

 

[idkkdi]

how do i solve
nC4 + nC5 =12C5

answers : n=11

notice that the q is literally an identity. (4.)

 

[CM_Tutor]

Yes, the direct solution here is to use the addition property:



Observing that the LHS in the given equation is



and matching it to the RHS of the addition property, we see that we have so as to also match and we have . Applying the addition property, this means we have that



So, equating this to the RHS of the equation we seek to solve, we get



and so

 

[CM_Tutor]

nC4 + nC5

= n! / [(4!) * (n-4)!] + n! / [ (5!) * (n-5)! ]

= ... = (n+1)! / [5! * (n-4)!]

[cut]

Cossine, you could have shortened the solution at this point by noting that



as the equation to solve would then become



from which the solution follows.