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avatar+359 

 

a) this is all i was given

 May 30, 2020
edited by TacoBell  May 30, 2020
edited by TacoBell  May 30, 2020
 #1
avatar+263 
-1

could you show what it looked like exactly?

 

if it's what you wrote right now then just combine like terms:

(mk-1)+(mk)

=mk-1+mk

=2mk-1

 

you didn't need the parentheses at the beginning because of the associative property i think.

 May 30, 2020
 #2
avatar+359 
-1

a) i will try 

b) no it is not 2mk-1

TacoBell  May 30, 2020
 #3
avatar+21556 
0

You are adding two combinations:  mCk-1 + mCk

 

1)  mCk  =  m! / [ k! · (m - k)! ]

2)  mCk-1  =  m! / [ (k - 1)! · ( m - (k - 1) )! ]  =   m! / [ (k - 1)! · ( m - k +1 )! ] 

 

To add them, they must have common denominators.

 

Formula 1) needs the term (m - k + 1) in its denominator because (m - k + 1)! = (m - k + 1) ·(m - k)!

 

Formula 2) needs the term k in its denominator because k! = k ·(k - 1)!

 

So:  1)  m! / [ k! · (m - k)! ] · [ (m - k + 1) / (m - k + 1) ] =  [ m!·(m - k + 1) ] / [ k! · (m - k + 1 )! ]

 

       2)  m! / [ (k - 1)! · ( m - k +1 )! ]  · [ k / k ]  =  [ m! · k ] / [ k! · (m - k + 1)! ]

 

Adding the numerators of these two parts:  [ m!·(m - k + 1) ] + [ m! · k ]  =  m! · [ (m - k + 1) + k ]

                       =  m! · (m + 1)

 

The full fraction:  [ m! · (m + 1) ] / [ k! · (m - k + 1)! ]  =  (m + 1)! / [ k! · (m - k + 1)! ] 

 

Probably, to make sense of this, try some particular problems, such as 8C3 + 8C2

 Jun 2, 2020
 #4
avatar
0

Just add 1 to the M of the last term and keep k the same: [m C k-1] + [m C k]=[m+1 C k].

 

Example: [10 C 4-1] + [10 C 4] =330 =[10+1 C 4] =11 C 4 =330.

 Jun 2, 2020

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