Find the number of ways in which: - (1) a selection (2) an

Question

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 · Answer

The solution is like this. 


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

wing · Answer

for permutation on 2 alike, 2 other different, why 4!/ 2!. I got lost at that stage. I understand all others though. can you explain further ?

Mayur · Answer

compare case 2 and caser 3. 
In case 2 you have to choose 2 alike from the 3 groups "O", "P", "R" and 2 others alike from the same group. Fro example, if you choose 2 "O"s, you have to choose 2 "P"s or 2 "R"s. So it is 4!/(2!*2!).

In case 3, you have to choose 2 alike as mentioned in case2 and 2 different from the 5 remaining letters. So, you have only 2 letters of a group and 2 letters in either of the following ways: -
1 letter from the remaining pair and 1 either from another pair or single letters (or)
2 letters from the single letters.

In either case, you have 2 similar letters and 2 differnt letters. Applying the permutations principle we get 4!/2!, because there is only 1 pair of the same letters in this case (3)

GoalStanford · Answer

Assume instead of two P's you have P1,P2.

As usual, you first get 4!. However P1 and P2 both are same.so P1P2 and P2P1 which you counted as two differrent ways,now you need to substitute as one PP.So devide by 2(=2!).

Similarly if youe have 3 P's , arrangements of P1P2P3, P1P3P2, P2P1P3, P2P3P1, P3P1P2, P3P2P1 are all actually same as PPP. So, you need to devide by 6 (=3!).

I would strongly recommend to practise on few more similar examples to get the concept even more clear.

Find the number of ways in which: - (1) a selection (2) an

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