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27. a) **In how many ways can a committee of 3 be chosen from an organization with 12 members?**

My answer: Number of Ways (12 3) = 220 ways

Correct

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b) **In how many ways can a president, vice-president, and treasurer be chosen from this same organization?**

My answer: C (12, 3) = 12! divided by 4! 3! = 3,326,400 different ways

Not sure how you got this. There are 12 choices for president, 11 for vice and 10 for secretary

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28. **a) How many 4-digit numbers can be formed from the set A = {0,1,2,3,4,5,6} if there is no repetition? (Note: 0123 is not a 4-digit # because it equals 123)**

My answer: Number of digits = (7 4) x 4! -1 ( minus one to get rid of the 0123 choice)

= 839 4 digit numbers can be formed

I think the point of the 0123 example was to say that no numbers with a 0 at the front would be considered as 4 digit numbers.

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b) **How many of the numbers in part a) are odd**

My answer: number of odd digits = 8 x 8 x 7 x 5 = 2240 odd digits This doesn't make sense because there can't be more odd digits than there are total choices in a)

Not sure what you are doing here. The 4-digit number will be odd if and only if the last digit is odd. How many of the 4-digit numbers will have an odd last digit?

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b) **How many of the numbers in part a) contain a 3? **

Find the number of 3-digit numbers you can make from the set {0,1,2,4,5,6}. Then multiply by the number of places you can add a 3 to those numbers. You will need to be careful with things like 3012 and 0312.