The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
+1 vote

Question How many different strings of length 2 can be made from the letters in 'ORONO'?

I am getting 7 strings as follows: OO, ON, OR, NO, RO, RN, NR
but given answer is 6.


asked in Combinatory by Veteran (43.9k points)
edited | 93 views

2 Answers

0 votes
Best answer

Actually the question is  different: 

How many different strings can be made from the letters in ORONO, using some or all of the letters?

Now, there are 3 strings of length 1  = 3  and

Seven strings of length two as given by @manu00x You are correct and book also gives correct answer

13 strings of length three = 3! using('O','R','N') + 3 using('O','O','N') + 3 using('O','O','R') + 1 using('O','O','O')

20 strings of length four = 12 using('O','O','N','R') + 4 using('O','O','O','N') + 4 using('O','O','O','R')

20 strings of length four = 20 using all characters

Total strings =63

I have also used formula Number of permutations of n elements when k1 are of type1 and k2 are of type2 and rest are different =

                                             $\frac{n!}{k_{1}! k_{2}!}$

answered by Boss (9.3k points)
yes, you are right! I didn't interpret the given solution correctly. Yes, I only posted that part of the question in which i had doubt.
0 votes

It should be 7

See here is 3 Os

So, here the trick lies

First from 3 Os, u r selecting 2 O's =$\binom{3}{2}\times \frac{2!}{3!}$=1..............................i

[Now why this 3!

Here is the reason

when there is 2 same letter we divide it by 2! ,

why this 2!, because say for $O_{1}O_{2}$

it can be arranged like $O_{1}O_{2} and O_{2}O_{1}$

That is why we divide by 2!

Now in case of 3!

there is$O_{1}O_{2}O_{3}$

Now we can select 2 among these 3 like

$O_{1}O_{2}$,$O_{1}O_{3}$,$O_{2}O_{1}$ ,$O_{2}O_{3}$,$O_{3}O_{1}$,$O_{3}O_{2}$

So 6 arrangement is possible

That is why we divide by 3!]

I guess upto this u r clear

Now it is easy

Select 1 O's among 3 O's and from N and R select 1 letter = $\binom{3}{1}\binom{2}{1}\frac{2!}{3}$=4 [divide 3 because we can select any one O's from 3 O's]...............................ii

and lastly select only N and R and arrange them=$\binom{2}{2}\times 2!$=2........................iii


Now adding i , ii and iii we get 7


answered by Veteran (83.7k points)

34,292 questions
41,038 answers
39,941 users