GATE CSE
First time here? Checkout the FAQ!
x
+1 vote
35 views

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 (14.9k points) 6 19 58 | 35 views

2 Answers

+1 vote
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 (5.8k points) 4 19 47
selected by
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.
+1 vote

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 (65.1k points) 35 222 625


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
Top Users Oct 2017
  1. Arjun

    23386 Points

  2. Bikram

    17068 Points

  3. Habibkhan

    8156 Points

  4. srestha

    6286 Points

  5. Debashish Deka

    5438 Points

  6. jothee

    4978 Points

  7. Sachin Mittal 1

    4772 Points

  8. joshi_nitish

    4344 Points

  9. sushmita

    3964 Points

  10. Rishi yadav

    3804 Points


Recent Badges

Reader #Rahul
Popular Question Arnabi
100 Club Rahul68
Popular Question Kanchan kumari
Old-Timer Santanu
Devoted Reader smsubham
Notable Question makhdoom ghaya
Good Comment Shivansh Gupta
Notable Question bhuv
Verified Human Sonas
27,316 questions
35,170 answers
84,072 comments
33,262 users