1

# Quick math

posted in Off Topic

i know i shouldnt be posting this on a vlr thread but rq question please if someone could do this;

How many different arrangements are there of the word ACTIVATED if the 2 A's must be AT LEAST 5 numbers away from each other

please ty ily if u can solve

im bad at math, but why tf would your teacher gave u this question?

its cambridge as maths questions this year im pre sure i fucked it up ;-;

i counted 5

Im smol brain sorry

f

it's been a while since i had permutations class, please correct me if i'm wrong
A C T I V A T E D
if 2 A's are separated by 5 letters then the word would be in the form: A (5 random letters) A (2 random letters)
If there were no repeated letters, the number of permutations would be 7! for 7 blank spaces
Since T is present twice, we divide it by 2!, which would mean number of permutations = 7!/2! = 2520

Similarly, the two A's can be separated by 6 and 7 letters too, total permutations = 2520 * 3 = 7560

forgive me if i am wrong i tried :)

you forgot to multiply with 3! on the 5 letter gap coz the 2 letters and the object can also be arranged
for 6 letter gap it should be multiplied with 2!

why would that make any difference, this is not combinations, order dosent matter, what matter is there are 7 blank spaces and 7 letters to fill it aka 7! arrangements

it depends on the method you use if you make the [A----A] an object you will have to arrange it, if you do it without object then it doesn't

a- - - - - a- -
a- - - - - - a -
a- - - - - - - a
-a- - - - - a -
-a- - - - - - a
— — a - - - - - a
the dashes are 7! for each of them, and there are 6 arrangements of ‘a’ , so 7! X 6. but there are two ‘t’ in activated, so divide the answer by 2 ( im not sure) so answer is 3 * 7!

activated has two Ts so its different

yeah, but the logic used here is rather than calculating specifically we find the ones that don't agree with our conditions and subtract them from all possible outcomes

but itll be too hard cus u need to consider A together, A 1 apart, A 2 apart, etc... wouldnt it be simpler to just calculate 5 6 7 ?

yeah you can do that but there's gotta be an easy way, I'll let you know if I find it

There is literally zero application for this question in the real world.

if you understand the basics it can be used in programming, this question is for strengthening basics

Well permutation combination has lots of applications in data management

ACTIVATED has 9 letters
we need to arrange 2 A's so not counting A, there are 7 letters and 7! ways to arrange those.
However the letter T is repeteated twice, so we divide 7!/2!
Question says atleast 5, so we multiple 7!/2! with 3 for all arangements for 5 letter gap, 6 letter gap and 7 letter gap

thus answer is 7!/2! * 3

I love how most serious answers in this thread come from India flags, y'all so smart

› check that that your post follows the forum rules and guidelines or get formatting help