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25 points, come on help me pls-example-1
User Thembi
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Answer:

There are 32 possible combinations when buying a home in this neighborhood.

Explanation:

We need to use a combination to find our answer. There are 5 options, and we need to find how many combinations can be made of some, all, or none.

  • If choosing none, there is 1 possible combination.
  • If choosing one, there are 5 possible combinations.
  • If choosing 2, there are 10 possible combinations.
  • If choosing 3, there are 10 possible combinations.
  • If choosing 4, there are 5 possible combinations.
  • If choosing all, there is 1 possible combination.

We use the permutation (combinations in no specific order) formula to find these: C(n,r)=n!/r!(n-r)! (sometimes C(n,r) is written as nCr, and in a calculator this is what you would see) , where n is the number of items to make a combination from (in this case n=5), and r is how many items are in the combination (in this case 0,1,2,3,4, or 5). Note that 0!=1.

  • None-5!/0!(5-0)! 5*4*3*2*1/1*5*4*3*2*1 = 1
  • One-5!/1!(5-1)! - 5*4*3*2*1/1*4*3*2*1 = 5
  • Two-5!/2!(5-2)! - 5*4*3*2*1/2*1*3*2*1 = 10
  • Three-5!/3!(5-3)! - 5*4*3*2*1/3*2*1*2*1 = 10
  • Four-5!/4!(5-4)! - 5*4*3*2*1/4*3*2*1*1 = 5
  • All-5!/5!(5-5)! - 5*4*3*2*1/5*4*3*2*1*1 = 1

Now we add our possible combinations: 1+5+10+10+5+1=32

So, there are 32 possible combinations when buying a home in this neighborhood.

User XAMlMAX
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