Exercise 29&30. Permutations with repititions and restrictions

Today we learned about Permutations again but looked at certain cases where there are either restrictions and repetition. We started off by learning a new formula {nPr=n!/(n-r)!} Where P is the number of arrangements, n= Number of items to chose from and r is the number of items being arranged. The first example was How many Permutation of there four of these letters are possible(A,W,L,S,O,D). first we look at our formula and we see that n is that number of variables to choose from which is 7 because there are 7 letters. Then you look for r, which is that number of variables being arranged and the question told us that it was 4. So we put this into our formula and solve. {7!/(7-4)!} Which can be reduced to {7!/(3)!}. Then you want to expand as 7!= 7654321 and 3! is 321. We can then cross out the common terms (321) and we are left with 7*6*5*4 Which is 840. That is our final answer.
The next kind of example we did was solving for Permutations with repetition. We looks at the formula {n!/j!k!m!} Which is NOT on our formula sheet. The first example was 'How many 5 diget numbers can be made from 4,6,1,6,4?' . At first we do our formula as we would for any other by putting n! on the top, then we look for the repetitions which are the "j!k!m!" Since 4 and 6 are repeated you but them as j! and k!. Since there are 2 of each digit you put 5!/2!2!. If there was 3 sixes instead of 2 you would have "5!/3!2!". Back to the original you are left with 54321/(2*1)(2*1). This can be reduced to 5*3*2*1 Which is 30.
The last type of permutations were examples with restrictions! The first example was "how many different arrangements could be made out of the word ALPHABET but it must start with the letter "B". Alphabet has 8 letters so you start just like you would any other problem. Since there is only ONE "b" you but a one in the first term such as 1_ _ _ _ _ _, leaving the next seven terms open. After you use the B your left with 7 variables and 6 after and so on resulting in 1 7 6 5 4 3 2 1. BUT since the "A" repeats you must put in under as your denominator as "2!" like you did in the last examples. Your left with {17654321/2*1}. The common terms cancel out and your left with 7*6*5*4*3, Which is 2520. Continue doing the exercises for more practice.