equinoxyy [#4]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 ;-;
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 :)
https://www.quora.com/Given-the-word-STAPLE-how-many-different-arrangements-are-there-of-the-letters-such-that-A-and-E-are-separated-by-at-least-one-letter
this might help you
note in your case both letters are the same so divide by 2!
M0K1 [#14]https://www.quora.com/Given-the-word-STAPLE-how-many-different-arrangements-are-there-of-the-letters-such-that-A-and-E-are-separated-by-at-least-one-letter
this might help you
note in your case both letters are the same so divide by 2!
activated has two Ts so its different
horse69 [#15]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
M0K1 [#16]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 ?
1phnxe [#17]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
HardStuckGold [#18]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
annoybrocc02 [#12]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! = 2520Similarly, 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!
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
M0K1 [#21]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
1phnxe [#17]There is literally zero application for this question in the real world.
Well permutation combination has lots of applications in data management
PRLV [#23]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
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