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# ???

-1
586
4

a) this is all i was given May 30, 2020
edited by TacoBell  May 30, 2020
edited by TacoBell  May 30, 2020

#1
-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
-2

a) i will try

b) no it is not 2mk-1

TacoBell  May 30, 2020
#3
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
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