If (1 1 !+2 2 !+3 3 !+ +n n !=11 !-1 ), then the maximum value of ( ^n C_r ) is

Step-by-step solution

( aligned & 1 1 !+2 2 !+ +n n ! & =(2-1) 1 !+(3-1) 2 !+ +[(n+1)-1] n ! & = 2 1 !+3 2 !+4 3 !+ (n+1) n ! & - 1 1 !+1 2 !+1 3 !+ +1 n ! & =(n+1) !-1 !=11 !-1 ! (given) aligned ) So, (n=10 ) Now, maximum value of ( ¹⁰ C_r ) occurs when (r= n 2 = 10 2 =5 ) Now, ( ¹⁰ C₅= 10 ! 5 !(10-5) ! =252 )