Hi ive got a question on permutations. This is probably a simple question but my mind is twisted right now so I can't seem to get b) and c) right. Not sure either if 4000 and 5000 is a typo for 40000 and 50000 respectively as the answers are small numbers??
a) How many different arrangements can be made from the numbers 3,4,4,5 and 6? ( which i figured was 5!/2! =60)
b) how many arrangements form numbers greater than 4000? (ans: 48)
c) how many form numbers less than 5000? (ans: 36)
Firstly I just wanna to assure you here that they are not typos, Im fairly certain that the questions are correct despite the answer does seem quite ridiculous. Good job on answering part a), remembering to divide by 2! which shows that you have recognised that there is a repetition.
Now for part B, I would like to invite you to draw just four underline dashes, each representing a number that can be placed in that spot. (i.e. __ __ __ __)
On the first underscore dash, there are only 4 numbers that can be placed there， which are 4, 4, 5, 6 (you cannot select 3 as the first digit because the number has to be greater than 4000)
. On the second underscore, there are also only 4 numbers that can be placed there (because a number, we dont know which one, has already been selected, leaving us with 4 options to choose from.)
On the third underscore, we will only have 3 options to choose from because we have already selected two of them from all the numbers that we are provided with. And finally on the last underscore we will only have 2 options to choose from because we have already selected 3 numbers from all the numbers that the question provides us with.
Hence, the total number of different arrangements will seem to be 4 x 4 x 3 x 2 = 96 ways. BUT, WE HAVE NOT YET ACCOUNTED FOR THE REPETITION OF 4 IN THE NUMBERS THAT WE ARE PROVIDED WITH. Hence similar to what you have done in part a), it is necessary for you to divide 96 by 2! and you will obtain an answer of 48.
Ok, now we are up to part c). This one will be extremely similar to part b). I would recommend you to follow what I did in part b) and attempt part c) yourself to check whether you have truly understood what I have done. I will still post the solution below, but have an attempt on yourself.
So, similar to part b), draw four underscores __ __ __ __, each representing a digit.
On the first digit, we can only have three options of numbers, because from all the numbers we are given, only three of them will suit the criteria of "a number that is less than 5000". These three numbers, evidently, are 3, 4, 4. The second digit will have 4 options for us to choose from, because there is a total of 5 numbers which we are provided with and we have already selected one. The third digit is the same, we will have 3 options to choose from because we have already chosen 2 numbers from the 5 for the first two digits. And for the final digit, we will have 2 options to choose from because we have already chosen 3 numbers for the first 3 digits.
Hence the total amount of numbers that we can make below 5000 will be (3 x 4 x 3 x 2) / 2! = 36 (2! is to account for the repetition of 4 in the numbers that we are provided with)
Anyways great questions, hope you have understood my solution. Don't hesitate to post more questions if you need further assistance!!!
Happy Physics Land