1. Select the Solved Problems. Explain the problem and present the solution also explaining (in detail)each step and why the solution is correct. 13! 7! 5.2. Compute: (a) (b) 10! 13! 13 . 12 . 11 . 10 . 9 . 8 . 7.6 .5-4.3 . 2. 1 (a) = 13 . 12 = 156. 11! 11 . 10 . 9 . 8 . 7 .6 .5 .4.3 .2 . 1 Alternatively, this could be solved as follows: 13! 13 . 12 . 11! II! = 13 . 12 = 156.1. Select the theorem explain the theorem in detail, provide additional examples that support the theorem. 5.6. Prove Theorem 5.3: (" + 1 ) = (, ")+ (") (The technique in this proof is similar to that of the preceding problem.) Now (, ",) + (" = 6-1).(4-r+1)! n! in! r! . (1 - r)! To obtain the same denominator in both fractions, multiply the first fraction by ; and the second fraction by , n-r+ 1 Hence ("1)+) = (n - r + 1) . n! r. (r - 1 ) ! . (n - r+1)! r!. (1-r+1). (1-r)! r . n! (n - r + 1 ) . n! ri(n - r + 1)! r!(m -r+ 1)! r . n! + (n - r+ 1).m! Ir+ (-r+ 1)] . n! r!(n - r + 1)! r!(n - r + 1)! (n + 1)n! (n + 1)! = r!(n - r+ 1)! r!(m - r+ 1)