1. In word problems, how can you tell whether you need to use the permutation formula or the combination formula?
2. How do you answer these kinds of questions:
a. How many 3-digit numbers can be formed from the set {0-9} if 0 cannot be the first digit, repetition is allowed and the number is divisible by 5? Answer: 180
b. How many 7-digit telephone numbers can be formed, assuming that no digit is used more than once and the first digit is not 0? Answer: 9 x 9 x 8 x 7 x 6 x 5 x 4 = 544, 320
My teacher gave us the answers but I really don't understand how she got them. I got Question B from my school book, which is about as explanatory as my teacher is. Help would be greatly appreciated!!! 1. It all depends on if the order of the arrangement is important or not. If it is, use the permutation formula; if it is not, then use the combination formula. A lot of times you have to think about the problem and decide if order is important. Here are a couple of examples:
E1: choosing 5 cards from a deck of 52 cards without replacement. In this case it doesn't matter what order the cards come in because the hand is still the same. That is if my five cards are As, Ac, Ad, Ah, 3d then I have "four aces" regardless of the order in which the cards were chosen.
E2: Out of 10 people in a club, 3 are to be selected to fill the positions of president, vice-president and treasurer. If A was chosen to be president, B for vice-president and C for treasurer that would be a distinct ordering. So order is important since having, say, B as president, C as VP and A as treasurer is a different make-up of the officer's of the club even though its still the same three people.
2a. What do you know?
i) 10 numbers (0 - 9)
ii) selecting a 3 digit number
iii) repetition is allowed
iv) first digit cannot be a 0. So only 9 numbers are possible for the first digit
v) number has to be divisible by 5, which means it has to end in a 0 or a 5.
So let's use all of this information to solve the problem
how many possibilities are there for the first digit? 9. Any number from 1, 2, ..., 9 is allowed.
how many possibilities for the 2nd digits? 10. now we can include 0 as well as 1, 2, ..., 9. Remember repetition is allowed.
how many possibilities for the 3rd digit? 2. Since the number has to be divisible by 5, the last digit has to be a 0 or a 5. So only 2 possibilities.
So, how many such 3 digit numbers are there that satisfy these requirements? The answer is found by multiplication.
9 * 10 * 2 = 180.
You have to multiply to account for all of the possible combinations. While this is a counting problem, it really doesn't fall into the "combination/permuation" category.
2b. What do you know?
i) need a 7 digit telephone number
ii) no digit is used more than once so repetition is NOT allowed
iii) first digit is not a 0 (so only 9 possibilities for the first digit)
To solve this we have to answer the following questions then multiply our answers together kind of like we did for part a.
Q1: How many numbers can we choose from for the first digit? The answer is 9 (0 is not allowed).
Q2: Once we have the first digit, how many numbers can we choose from for the 2nd? This is a little tricky. Even though we can now use 0, we can't say 10 because repetition is not allowed. That is, whatever the first number was, it can't be used again. So we really only have 9. Why? We have 8 possibilities from 1, 2, ..., 9 (remember we can't have repetitions) plus 0 is also a possibility.
Q3. Once we chose the first 2 digits, how many numbers can we choose from to get the 3rd digit? since we can't have repetition, there is only 8 possible numbers left for us to choose from.
The pattern continues.
Q4: How many for the 4th digit once we have selected the first 3? 7
Q5. How many for the 5th digit once we have selected the first 4? 6
Q6: How many for the 6th digit once we have selected the first 5? 5
Q6: How many for the 7th digit once we have selected the first 6? 4
So the total number of ways of creating a 7 digit telephone number when the first number cannot be a 0 and no repetition is allowed is:
9 * 9 * 8 * 7 * 6 * 5 * 4 = 544,320
Notice that once we select the first digit, the number of ways of choosing the remaining digits is a permutation 9P6 - choosing 6 digits from 9 when order is important. For a telephone number order is important. The telphone number 1234567 is different than 7654321. 9P6 = 9!/(9 - 6!) = 9!/3!. If 0 was allowed for the first digit then the number of ways of selecting a 7 digit telephone number without replacement (ie repetition is not allowed) when order is important would be 10P7.
I hope this helps. a. just add 5 from 100 til it gets to 995
b. i dunno Combination problems are those where order is not importnt, permutations the order items are selected is important. |
