Let S = _ k=0 ⁹ 1 k!(12-k)!(k+3)!(9-k)! . If S can be expressed in the form 24! a! , b! , (c!)^2 , where a, b, c are positive integers and a < b , then the value of 2b - a - c is equal to

Step-by-step solution

S = _ k=0 ⁹ 1 k!(12-k)!(k+3)!(9-k)! Multiply and divide the expression by (12!)^2 : S = 1 (12!)^2 _ k=0 ⁹ ( 12! k!(12-k)! ) ( 12! (k+3)!(9-k)! ) S = 1 (12!)^2 _ k=0 ⁹ ¹²C_ k ¹²C_ k+3 Using the property ^ n C_ r = ^ n C_ n-r , we can write ¹²C_ k+3 = ¹²C_ 9-k . S = 1 (12!)^2 _ k=0 ⁹ ¹²C_ k ¹²C_ 9-k The sum _ k=0 ⁹ ¹²C_ k ¹²C_ 9-k represents the coefficient of x^9 in the expansion of (1+x)¹²(1+x)¹² = (1+x)²⁴ . By Vandermonde's convolution, this sum is equal to ²⁴C₉ = 24! 9!15! . Substituting this back into S : S = 1