## mathmath333 one year ago counting question

1. 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}}

2. rational

How many vowels are there ?

3. mathmath333

all vowels are there

4. rational

put them in a bag and call it $$\phi$$

5. rational

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

6. mathmath333

=(8/2)!

7. anonymous

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

8. rational

you mean 8!/2 ?

9. mathmath333

ya this one

10. rational

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

11. mathmath333

can vowels be repeated

12. rational

what letters do you have in the bag ?

13. mathmath333

aeiou

14. rational

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

15. mathmath333

5!

16. rational

Yes, so the final answer is ?

17. mathmath333

8!*5!/2

18. rational

looks good!

19. mathmath333

ok the next one

20. ikram002p

separated than the first condition ?

21. mathmath333

P and S are at distance of 4 places always

22. rational

First find the number of ways of placing $$P, S$$ : |dw:1439069671413:dw|

23. mathmath333

12*11

24. rational

nope, try again

25. mathmath333

11

26. rational

|dw:1439070002368:dw|

27. rational

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

28. mathmath333

6

29. rational

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

30. 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 ?

31. mathmath333

7

32. rational

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

33. rational

Next, arrange remaining 10 letters

34. mathmath333

10!/2

35. rational

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

36. 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 .

37. rational

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

38. mathmath333

ohk i see the difference :D

39. mathmath333

hmm this little silly doubts make my brain spin

40. rational

its natural, my brain spins too :)

41. rational

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

42. mathmath333

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

43. 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

44. mathmath333

yes thumbs up!

45. mathmath333

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