Fifth Day!

Hey guys, this is Rachel! So, today we had another quiz for solving "n". After the quiz, Mr. P made a brief explanation for both quizzes, (there were 2 different ones) to the whole class. Then he continued our lesson to "Permutations with Repetitions and Restrictions" and "Permutations with Case Restrictions"  


Permutations with Repetitions and Restrictions AHEM *make sure you know this because this is apparently popular on tests and exams* :) Well, first of all, repetitions means you can have repeats. So,you may use numbers more than once when permuting. If we are working with repetitions we are NOT able to use the formula. Just a little reminder. :P Examples are in our notes. I am just explaining how we got answer our answers on some.

Ex #2 Consider the digits 2, 4, 5, 6 and 8. If repetitions are allowed, find...
a) How many four digit numbers can be formed?
Well, since there are 5 numbers in total and repetitions are allowed, we are able to have, =5x5x5x5 or 5^4 ("4" represents the digits needed) =625

Ex #3 a) How many four digit numbers can be formed? (this time, no repetitions are allowed) Since there are 5 different numbers and there are only 4 digits available, we must think in 5P4. ("4" represents the digits needed). =5x4x3x2 =120 We can also use the formula because REPETITIONS ARE NOT allowed. =5!/(5-4)! =120

Ex #4 A family consisting of 7 of the parents and 5 kids are going to be arranged in the photo. Calculate the number of permutations if all 5 children must be seated together. Since all the children must sit together, they are known to be 1 entity. They can also, be moved around even though they are in one group. for example, k= kid. k1 k2 k3 k4 k5 but can be k2 k3 k4 k5 k1 and all the other permutations. since there are 5 children they will represent as 5! Since the mother and father can be in any place without disrupting the group of their kids they are known to be 2 entities or 2!. Since the group of the kids are 1 entity they are represented as 1!. So the total of entities is equal to 3! =5!x3! =720  

Permutations with Case Restrictions
Ex #1 c) How many 6 letter "words" are possible using the letters in GADGET? Since there are 6 letters in the word and we need 6 letters to make up "words" we can represent this as 6P6 or 6!. But make sure to keep in mind that there are 2 G's in this word, meaning we must divide by 2. =(6!)/2 =720/2 =360

Ex #2 Three sets of books are being arrancged on the shelf. The first set has 5 volumes, the second set has 3, and the third set has 2. In how many ways can the books be arranged if the volumes of each set are to be kept together? Think of sets. first set = (5!) second set =(3!) third set = (2!) *don't forget that they are 3 different entities so (3!) When we put all those together it comes to, =(5!)(3!)(2!)(3!) =8640 *To end this, just make sure to read the word problems carefully. Because it's very easy to screw up a little thing. LOL* BYE Y'ALL :D