Combination question

  • Thread starter Bachelier
  • Start date
  • #1
376
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How many sequences of zeros and ones of length 7 contain exactly 4 ones and 3 zeros?

I'm having a hard time figuring out why it is 7 Chooses 4?

thx
 

Answers and Replies

  • #2
662
1
Select any four spaces within the seven spaces in 7C4 ways, and put 1's in each of the four spaces. Where can all the zeros go now? Notice the identity 7C4=7C(7-4)=7C3. Can
you see, in the context of this problem , why 7C4=7C3?
 
  • #3
376
0
Can
you see, in the context of this problem , why 7C4=7C3?
Yes, because we for 3 zeros and 4 ones.
 
  • #4
662
1
Exactly. Perfect. This is the general identity nCk =nC(n-k).
 
  • #5
376
0
Exactly. Perfect. This is the general identity nCk =nC(n-k).
Thank you :)
 
  • #6
662
1
No problem; glad to help.
 
  • #7
chiro
Science Advisor
4,790
132
How many sequences of zeros and ones of length 7 contain exactly 4 ones and 3 zeros?

I'm having a hard time figuring out why it is 7 Chooses 4?

thx
If you want to get some intuition, what you can do is to start with 1111000, and then find every way of re-arranging these digits so that each rearrangement is different from each other. For example 1111000 goes to 1110100, 1110010, 1110001 and so on.

Using this as your intuition base, you can develop a factorial relationship which will eventually give you the "N choose R" formula that you use.
 

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