counting question

- mathmath333

counting question

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- mathmath333

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

- rational

How many vowels are there ?

- mathmath333

all vowels are there

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## More answers

- rational

put them in a bag and call it \(\phi\)

- rational

how many ways can arrange the objects \(\{\phi, ~P, ~R, ~M,~T,~T,~N,~S\}\) ?

- mathmath333

=(8/2)!

- anonymous

12!/(12-5)! for i?

- rational

you mean 8!/2 ?

- mathmath333

ya this one

- rational

Yes, next unpack the bag, how many ways can you arrange 5 vowels ?

- mathmath333

can vowels be repeated

- rational

what letters do you have in the bag ?

- mathmath333

aeiou

- rational

they all are distinct, how many ways can you arrange them ?

- mathmath333

5!

- rational

Yes, so the final answer is ?

- mathmath333

8!*5!/2

- rational

looks good!

- mathmath333

ok the next one

- ikram002p

separated than the first condition ?

- mathmath333

P and S are at distance of 4 places always

- rational

First find the number of ways of placing \(P, S\) :
|dw:1439069671413:dw|

- mathmath333

12*11

- rational

nope, try again

- mathmath333

11

- rational

|dw:1439070002368:dw|

- rational

If one letter of {P, S} is at index 1, where can the other letter be ?

- mathmath333

6

- rational

Yes, {P, S} can be mapped to {1, 6}

- rational

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 ?

- mathmath333

7

- rational

and you can swap the locations, so there are 7*2 = 14 different ways to arrange {P, S}

- rational

Next, arrange remaining 10 letters

- mathmath333

10!/2

- rational

Yes, so total num of permutations = 14(10!/2)

- mathmath333

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 .

- rational

if the restriction is not there, then there are 12!/2 different permutations
not 10!/2

- mathmath333

ohk i see the difference :D

- mathmath333

hmm this little silly doubts make my brain spin

- rational

its natural, my brain spins too :)

- rational

listing the possibilities and looking at the problem from alternative ways usually helps...

- mathmath333

but in the way of learning its essential to get rid of the small doubts

- rational

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

- mathmath333

yes thumbs up!

- mathmath333

remembering school days where teachers used to run after students after student asked questions

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