counting question

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\(\large \color{black}{\begin{align} & \normalsize \text{In how many ways can the letters of the word PERMUTATIONS be arranged if the}\hspace{.33em}\\~\\ & \normalsize \text{ (i) vowels are all together }\hspace{.33em}\\~\\ & \normalsize \text{ (ii) there are always 4 letters between P and S.}\hspace{.33em}\\~\\ \end{align}}\)
How many vowels are there ?
all vowels are there

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Other answers:

put them in a bag and call it \(\phi\)
how many ways can arrange the objects \(\{\phi, ~P, ~R, ~M,~T,~T,~N,~S\}\) ?
=(8/2)!
12!/(12-5)! for i?
you mean 8!/2 ?
ya this one
Yes, next unpack the bag, how many ways can you arrange 5 vowels ?
can vowels be repeated
what letters do you have in the bag ?
aeiou
they all are distinct, how many ways can you arrange them ?
5!
Yes, so the final answer is ?
8!*5!/2
looks good!
ok the next one
separated than the first condition ?
P and S are at distance of 4 places always
First find the number of ways of placing \(P, S\) : |dw:1439069671413:dw|
12*11
nope, try again
11
|dw:1439070002368:dw|
If one letter of {P, S} is at index 1, where can the other letter be ?
6
Yes, {P, S} can be mapped to {1, 6}
By sliding to right, you can see that {P, S} can take on locations : {1, 6} {2, 7} {3, 8} {4, 9} {5, 10} {6, 11} {7, 12} how many are they ?
7
and you can swap the locations, so there are 7*2 = 14 different ways to arrange {P, S}
Next, arrange remaining 10 letters
10!/2
Yes, so total num of permutations = 14(10!/2)
if the restriction of 4 places is not there then their are 10!/2 ways and the restrictions are their then there are 14*10!/2 ways but as the restrictions are more the the number of ways should decrease .
if the restriction is not there, then there are 12!/2 different permutations not 10!/2
ohk i see the difference :D
hmm this little silly doubts make my brain spin
its natural, my brain spins too :)
listing the possibilities and looking at the problem from alternative ways usually helps...
but in the way of learning its essential to get rid of the small doubts
haha! its good to ask questions there is an old saying : "there are no stupid questions, only stupid answers" remember that when asking a question which you feel is small/silly ;p
yes thumbs up!
remembering school days where teachers used to run after students after student asked questions

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