Permutations & Combinations

[m_mayurprabhu]

Find the number of ways in which: -
(1) a selection
(2) an arrangement,
of 4 letters can be made from the letters of the word PROPORTION.

------------
Mayur

[TheAkshaT]

Originally Posted by m_mayurprabhu

Find the number of ways in which: -
(1) a selection
(2) an arrangement,
of 4 letters can be made from the letters of the word PROPORTION.

First of all as the question says 4 letter and NOT 4 difft letters....so

For Selection...

10C4 Edit : this is wrong

For arrangement...

4!*(6C4*1 + 5C1*1/3! + 3C1*5C2*1/2! + 3C2*1/2!2! )

-Akshat

[voodoochild]

Originally Posted by TheAkshaT

For Selection...

10C4

proportion

I guess it would be 10C4/(2!*2!*3!)

Vishal

[TheAkshaT]

Originally Posted by voodoochild

proportion

I guess it would be 10C4/(2!*2!*3!)

infact it'll be (6C4 + 5C1 + 3C1*5C2 + 3C2)

-Akshat

[m_mayurprabhu]

Ok guys - You've put in your efforts. Let me give the solution.

There are 10 letters of 6 different sorts, namely O, O, O; P, P; R, R; T; I; O; N

In finding groups of 4, these may be classified as follows: -
(1) 3 Alike, 1 Different
(2) 2 Alike, 2 Others Alike
(3) 2 Alike, the other 2 Different
(4) All 4 different.


(1) Each of the 5 letters, (P, R, T, N) can be taken with the single group of the 3 like letters “O”. So, the selection can be made in 5 Ways.

(2) We have to choose 2 out of the 3 pairs O, P, R. So, the selection can be made in 3C2 ways. = 3 ways.

(3) We select 1 of the 3 pairs and then the 2 from the remaining 5 letters. This gives 3C2 X 5C2 ways = 30 Ways.

(4) All the 4 can be selected in 6C4 ways = 15 (We have to take 4 different letters from O, P, R, T, I, N)

Thus, the total number of ways = 5+3+30+15 = 53
Arrangements: - We have to permute in all possible ways each of the 4 groups above.

(1) 5 X 4! / 3! = 20 arrangements
(2) 3 X 4! / (2!X2!) = 18 arrangements
(3) 30 X 4! / 2! = 360 arrangements
(4) 15 X 4! = 360 arrangements

Total = 20+18+360+360 = 758 arrangements.

-----------
Mayur

[TheAkshaT]

Originally Posted by m_mayurprabhu

Thus, the total number of ways = 5+3+30+15 = 53
Total = 20+18+360+360 = 758 arrangements.

Errr....

My solutions give same ans!

-Akshat

[m_mayurprabhu]

Akshat,
Actually I did not see your II posting regarding selection. Henceforth, please put the answer and don't just leave it like that. You have done a good job

[raakesh_natraj]

guys..
perm and comb is really hurtin me a lot...can u guys help out..jus a coupla querries..

1) how can n identical obj be distbtd amoung 'p' persons where each person can get any number of lots...
2) how can 'n' identical obj be distbtd amoung 'p' persons when each gets atleast one lot....

do explain ur answers clearly guys so that ppl like me can get salvation from the evils of perm an comb...
thnx.....rak

[RahulSharma]

I suggest you to look for the answer in the Algebra book Hall and Kinight, or ask this from a Math faculty in any institute.

Not easy to explain with given symbols on the keyboard.

The answer is

(1) (p+n-1)C(p-1) where aCb means a!/(b!(a-b)!)

(2) (n-1)C(p-1)

[cattie]

Ghulam! r u sure of the ans?
Well, have a look at mine.

(1) when each person can get any number of the lot, the 1st obj can be given to any of the 'p' persons, similarly the 2nd obj, 3rd obj .... and so on. So the total number of ways possible is p*p*p*.....(n times)=p^n
Now since all the objs r identical so the number of ways get multiplied by the factor of 1/n!
Hence the ans comes out to be (p^n)/n!.

(2) Here since each of them has to be given atleast one, lets first distribute one to each one of the 'p' persons. This can be done in only one way as all the objs r similar.
so now we r left with (n-p) objs. these (n-p) objs can be ditributed in
pC(n-p) ways.
Hence the ans to the 2nd one is pC(n-p).

NOTE: pC(n-p) could also be written as pC(n-p)*(n-p)!/(n-p)!=pC(n-p).

Let me know if I'm wrong.