Reducing combinations in fractions b ∗ ( b − 1 k − 1 ) ( r n − k ) n ∗ ( b k ) ∗ ( r + b n ) ( r n − k ) ( b k ) How would I go about reducing this into k n ?My attempt was b ! n ∗ ( k − 1 ) ! ( b − k − 2 ) ! ∗ k ! b ! = k ! n ∗ ( k − 1 ) ! ( b − k − 2 ) ! = k n ∗ ( b − k − 2 ) ! Don't know where the errors are / where to go from here

Question

Reducing combinations in fractions            b      ∗                                                  (                                                          b              −              1                                      k              −              1                                                          )                                                                                      (                                            r                          n              −              k                                                          )                                                  n      ∗                                                  (                                            b            k                                              )                                            ∗                                          (                                                r            +            b                    n                                      )                                                                                  (                                            r                          n              −              k                                                          )                                                                                      (                                            b            k                                              )                                          How would I go about reducing this into       k    n  ?My attempt was             b      !              n      ∗      (      k      −      1      )      !      (      b      −      k      −      2      )      !        ∗            k      !              b      !        =            k      !              n      ∗      (      k      −      1      )      !      (      b      −      k      −      2      )      !        =      k          n      ∗      (      b      −      k      −      2      )      !      Don&#039;t know where the errors are / where to go from here

lesinetzgl5 · Accepted Answer

b                                                  (                                                          b              −              1                                      k              −              1                                                          )                                                  n                                                  (                                            b            k                                              )                                            =            b      ⋅                        (          b          −          1          )          !                          (          k          −          1          )          !          (          b          −          k          )          !                            n      ⋅                        b          !                          k          !          (          b          −          k          )          !                      =      k    n  because   b  ⋅  (  b  −  1  )  !  =  b  ! and   k  !  =  k  (  k  −  1  )  !

Reducing combinations in fractions (b((b-1),(k-1))((r),(n-k)))/(n((b),(k)))*(((r+b), (n)))/(((r), (n-k))((b),(k)) How would I go about reducing this into k/n?

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