Log in Anne 9 years agoPosted 9 years ago. Direct link to Anne's post “Can someone clarify why 0...” Can someone clarify why 0! = 1? I'm really getting lost on the last paragraph, specifically: "It's the product of all integers greater than or equal to 1 and less than or equal to 0. But there are no such integers. Therefore, we define 0! to equal the identity for multiplication, which is 1." • (33 votes) Cameron 9 years agoPosted 9 years ago. Direct link to Cameron's post “n! is the product of all ...” n! is the product of all integers greater than or equal to 1 and less than or equal to n so when n=0 we have Hope this makes sense (81 votes) Maddie Walker 6 years agoPosted 6 years ago. Direct link to Maddie Walker's post “I'm still really confused...” I'm still really confused on this subject. If one of y'all could explain how n and n! work in a really simple almost kindergarten version, that would be amazing! I'm trying to relate this so something that I have already learned/know about so the concept will be easily identified and understandable. Thanks in advance! • (9 votes) Cameron 6 years agoPosted 6 years ago. Direct link to Cameron's post “n! is the product (multip...” n! is the product (multiplication) of all the whole numbers from n to 1 0! is a special case. We say that 0! = 1. We use factorials when we look at permutations and combinations. If we have n different items, then we can permute (arrange) them in n! different arrangements. If we have n items there are n!/(k! * (n-k)!) different combinations of choosing k items from those n items. For more info check out: (24 votes) lauren.finkle 9 years agoPosted 9 years ago. Direct link to lauren.finkle's post “In the for loop when we s...” In the for loop when we say result = result * i; how do these two results not override each other? How does the computer know that one result is referring to the previous sum of the multiplications and the other is that value times i? • (2 votes) jdsutton 9 years agoPosted 9 years ago. Direct link to jdsutton's post “`result = result * i;` is...” When you use the (12 votes) Vu Tran 9 years agoPosted 9 years ago. Direct link to Vu Tran's post “In the example given abov...” In the example given above where you have n t-shirts and want to be able to pack only k amount, the given formula is n!/(k!⋅(n−k)!) In the permutations course given in the link, the formula is stated to to be n!/(n-k)! Can someone explain where the extra k! in the denominator came from? • (3 votes) Cameron 9 years agoPosted 9 years ago. Direct link to Cameron's post “n!/(k! (n-k)!) would tell...” n!/(k! (n-k)!) would tell you the number of ways you could pick k t-shirts from n shirts where the order does not matter n!/(n-k)! would tell you the number of ways you could pick k t-shirts from n shirts where the order matters e.g. the first shirt you pick you write 1 on it, the 2nd shirt you write 2 on it, etc. Dividing by k! accounts for the number of ways each selection of k shirts will be counted. (8 votes) Barbara DiLucchio 5 years agoPosted 5 years ago. Direct link to Barbara DiLucchio's post “when you are writing pyth...” when you are writing python code for any given factorial say one that a user enters like 11. what if you also need to determine which values are odd and which are even is it true that they are always even? do you ever get odd results going down thru the range values? • (3 votes) Jesse 5 years agoPosted 5 years ago. Direct link to Jesse's post “All the results of the fa...” All the results of the factorial function are always even UNLESS it's 1! or 0!. Therefore for all n! such that n doesn't equal 0 or 1 the result is even. That's because 0! and 1! equals 1, which is odd. This is because an odd number multiplied by an even number is even, and an even number multiplied by an even number is also even. Therefore for every n! where n > 1 an even number is involved, making the answer even. Hope that helped! (3 votes) Renan Teixeira Ferraz 5 years agoPosted 5 years ago. Direct link to Renan Teixeira Ferraz's post “The autograder isn't work...” The autograder isn't working for me. The console is showing me that the code works, but the autograder doesn't. • (3 votes) Jesse 5 years agoPosted 5 years ago. Direct link to Jesse's post “That happened to me too. ...” That happened to me too. I reported it and am hoping they'll fix the issue. (the console showed it worked, and my code is just like the code in the hint). (2 votes) cmb717 3 years agoPosted 3 years ago. Direct link to cmb717's post “I can not get the next ch...” I can not get the next challenge completed what am i doing wrong? Here is my code (edited to only show relevant parts) for(var i= result; i>=n; i--) { • (2 votes) Cameron 3 years agoPosted 3 years ago. Direct link to Cameron's post “Let's review some for loo...” Let's review some for loop basics: A simple for loop that would print out the numbers from 1 to 10 would be: Many find the BASIC version of this a bit easier to understand: There are a couple of things that you almost never want to do inside the body of the for loop: Doing either of those usually leads to unpredictable behaviour. Here's some extra hints using the hint code: //think about calculating n! by hand //we need to update the value of result here, so that it //changing the counter i, or the value that it is going to, would be bad here //we are returning the value of result here Good Luck (3 votes) Matthew Narodowg 3 years agoPosted 3 years ago. Direct link to Matthew Narodowg's post “my code keeps coming up w...” my code keeps coming up with this Hm, are you hard-coding the number to count up to? • (2 votes) Cameron 3 years agoPosted 3 years ago. Direct link to Cameron's post “If you put a number in th...” If you put a number in the condition of the for loop you'll get that. e.g. If you put a number like 5 instead of using a variable, that is what is called "hard coding". Hard coding makes your program work for only one specific input. We want our programs to be flexible and work with different inputs, so we use variables. (2 votes) SecretCoder 2 years agoPosted 2 years ago. Direct link to SecretCoder's post “Why do they have a '!' af...” Why do they have a '!' after numbers like: 0!, 1!, n! • (2 votes) Cameron 2 years agoPosted 2 years ago. Direct link to Cameron's post “From the first paragraph ...” From the first paragraph of the article: (2 votes) Eden Washington 2 years agoPosted 2 years ago. Direct link to Eden Washington's post “I can't get this complete...” I can't get this completed.. I've done the coding statements as I knew how to do But I kept on getting • (2 votes) Cameron 2 years agoPosted 2 years ago. Direct link to Cameron's post “The above calculation `re...” The above calculation (2 votes)Want to join the conversation?
or more simply:
the product of all integers from 1 to n (starting at 1 and going up to n)
i.e. 1 * 2 * 3 * 4 * 5 * ... * n-1 * n
the product of all integers from 1 to 0 (starting at 1 and going up to 0)
But 1 is already larger than 0 so there are no integers from 1 to 0 (counting up)
So what do we use for the product of no integers ?
We use 1.
Why ?
Because it is convenient, it makes makes math easier, and most importantly it makes our formulas that could calculate 0! make sense. The article gives an example of how it makes our formulas make sense. It shows that when we calculate the number of ways of choosing a group of n items from n items (which should obviously be 1), the formula needs 0! to be 1.
e.g.
3! = 3 * 2 * 1
2! = 2 * 1
1! = 1
Permutations tell us how many different ways we can arrange things if their order matters.
Combinations tells us how many ways we can choose k item from n items if their order does not matter.
e.g.
If we have 3 items: A,B,C then there are 3! = 3 * 2 * 1 = 6 different permuations
Here they are:
ABC, ACB, BAC, BCA, CAB, CBA
e.g.
If we 5 items: A,B,C,D,E and we want to choose 2 items at a time then there are:
5!/( 2! * (5-2)!) = 5! / (2! * 3!) = 120/(2 * 6) = 10 different combinations
Here they are:
AB, AC, AD, AE, BC, BD, BE, CD, CE, DE
Note: CB would be considered as the same combination as BC, because order doesn't matter
https://www.khanacademy.org/math/precalculus/prob-comb/combinatorics-precalc/v/factorial-and-counting-seat-arrangements
https://www.khanacademy.org/math/precalculus/prob-comb/combinations/v/introduction-to-combinationsresult = result * i; is really telling the computer to do this:
1. Compute the value of result * i and store it somewhere
2. Take the stored value and assign result to that value= operator, the entire right hand side is evaluated to some value, and then that value is put into the left hand side variable. Everything is happening in steps, not all at once.
e.g. Suppose k= 3 and one possible selection of shirts is shirts: 5, 7 and 13
You could have picked those shirts in these 3! = 6 different ways:
5 then 7 then 13 or,
5 then 13 then 7 or,
7 then 5 then 13 or,
7 then 13 then 5 or,
13 then 5 then 7 or,
13 then 7 then 5var result = 1;
n=i*n;
}
I looked through the comments for help but all I could find were Cameron's cryptic answers can I have a more firm hint please. (not saying Cameron did a bad job just saying I can not understand their hints)//This for loop will:
//-begin the counter at start,
//-do some stuff each loop
//-then increment counter
//until the counter exceeds last
for( var counter = start; counter <= last; counter++){
//this is the body where we do some stuff
}for( var i = 1; i <= 10; counter++){
println(i);
}FOR I=1 TO 10
PRINT(I)
NEXT I
1) change the value of the counter
2) change the value of the number you are stepping the counter towards e.g. last in the first examplevar factorial = function(n) {
//this will set the value of result before we start looping
var result = ;
// n! = 1 * 2 * 3 * ... * n-2 * n-1 * n
// what number do we start at ?
// what number do we end at ?
for(var i = ; i <= ; ) {
//will be equal to n! by the time we stop looping
;
}
//so result should now hold the value of n!
return result;
};
message
idk what i'm doing wrongi <= 5
"We indicate the factorial of n by n!. It's just the product of the integers 1 through n. For example, 5! equals 1 * 2 * 3 * 4 * 5 or 120. (Note: Wherever we're talking about the factorial function, all exclamation points refer to the factorial function and are not for emphasis.)"
"Hm, do all your assertions pass?" for this code I was doing, what do i need to add or fix?var result = 1;
for(var i= 0; i<=n; i++) {
result=(n+1)*(i+1);
}
return result;
};result=(n+1)*(i+1); is incorrect.
A few things to remember:
n! = 1 * 2 * 3 * ... * n
and
n! = n * (n-1)!
and
0! = 1 (which is same as result before the for loop. So, if it skipped the calculations in the for loop it would be nice)
