Factorial fact!

Algebra Level 3

( 2 n ) ! ( n 3 ) ! = 72 ( 3 n 7 ) ! (2n)!(n-3)!=72(3n-7)!

Find the value of n n satisfying the equation above.


The answer is 6.

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2 solutions

Achraf Wassim
Jul 18, 2019

( 2 n ) ! × 3 × ( n 2 ) × ( n 3 ) ! = 72 × ( 3 n 6 ) × ( 3 n 7 ) ! (2n)!×3×(n-2)×(n-3)!=72×(3n-6)×(3n-7)!

Multiplying both sides by ( 3 n 6 ) (3n-6) , where 3 n 6 = 3 × ( n 2 ) 3n-6=3×(n-2)

( 2 n ) ! × 3 × ( n 2 ) × ( n 3 ) ! = 72 × ( 3 n 6 ) × ( 3 n 7 ) ! (2n)!×3×(n-2)×(n-3)!=72×(3n-6)×(3n-7)!

3 × ( 2 n ) ! × ( n 2 ) ! = 72 × ( 3 n 6 ) ! 3×(2n)!×(n-2)!=72×(3n-6)!

( 2 n ) ! × ( n 2 ) ! = 24 × ( 3 n 6 ) ! (2n)!×(n-2)!=24×(3n-6)!

Since, 24 = 1 × 2 × 3 × 4 = 4 ! 24=1×2×3×4=4!

( 2 n ) ! × ( n 2 ) ! = 4 ! × ( 3 n 6 ) ! (2n)!×(n-2)!=4!×(3n-6)!

( 2 n ) ! 4 ! = ( 3 n 6 ) ! ( n 2 ) ! \frac{(2n)!}{4!}=\frac{(3n-6)!}{(n-2)!}

P ( 2 n , 2 n 4 ) = P ( 3 n 6 , 2 n 4 ) P(2n,2n-4)=P(3n-6,2n-4)

3 n 6 = 2 n n = 6 3n-6=2n⟹n=6

Roger Erisman
Jul 11, 2019

Notice that any n less than 3 creates a factorial of a negative number.

Subtract the right side so the expression equals 0.

Beginning with 3 try integers:

Integer. Result.

  1. 576

  2. 31680

  3. 4354560 It seems to be increasing but try one more

  4. O. Eureka!

You have only demonstrated that n=6 works. But is there any other solution that works?

Pi Han Goh - 1 year, 11 months ago

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