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11 Cards in this Set
- Front
- Back
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The permutation of k-subsets of an n-element set is: |
P(n, k) = k! * c(n, k) |
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How many 3-digit numbers can be formed from the {1, 2, 3, ..., 9}? |
The number of 3 subsets of a 9-set is c(9,3) = 9!/3!6! * 3! = 9!/6! = 504 |
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What is the meaning of P(n, k)? |
The number of possible sequences that can be formed from c(n, k) |
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What is probability? |
Let E be a subset of S. Then the probability of E, P(E), is defined as P(E) = n(E)/n(S) |
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What is Pascal's Identity? |
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What is the sum rule of C(n, k)? |
C(n, 0) + C(n, 1) + C(n, 2) + ... + C(n, n) = 2^n |
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What is the nonempty subset of a 10-set? |
There are 2^n subsets of an n-set, therefore there are 2^10 subsets of a 10 set. There are C(10, 0) subsets of the 10 set that are empty. Hence there are 2^10 - C(10, 0) empty sets or 2^10 - 1 nonempty sets. |
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How many nonempty subsets of {1, 2, ..., 100} are there? |
C(100, 1) + C(100, 2) + C(100, 3) + ... +C(100, 100) = 2^100 - C(100, 0) = 2^100 - 1 |
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How many ways are there to choose 3 or more people from 100? |
There are 2^100 ways to choose a subset of 11 people. However, the subsets with 0, 1 and 2 people are not allowed. So the number of of subsets with 3 or more people are 2^100 - C(100, 2) - C(100, 1) - C(100, 0) |
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Proof of sum rule |
Since (1+x)^n = C(n, 0)x^0 + C(n, 1)x^1+ ... + C(n, n)x^n To get 2^n we must let x = 1 --> 2^n = C(n,0) + C(n,1) + ... + C(n, n) |
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P(n, k) represents... |
the number of n things taken r at a time |
