### Discussion :: Permutation and Combination - General Questions (Q.No.3)

3. | In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? |
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Answer: Option D Explanation: In the word 'CORPORATION', we treat the vowels OOAIO as one letter. Thus, we have CRPRTN (OOAIO). This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
Required number of ways = (2520 x 20) = 50400. Video Explanation: https://youtu.be/o3fwMoB0duw |

Pritesh Patel said: (Sep 23, 2010) | |

Why we dividing 7! and 5! by 2! and 3!, respectively. Is there any formula for that? What is the Logic Behind it? Thanks. |

Satty said: (Oct 12, 2010) | |

I have the same question, Is there a formula behind ? |

Priyanka said: (Oct 16, 2010) | |

I also can't understand division. Is there any formula ? |

Nishtha Sharma said: (Dec 1, 2010) | |

Here we can simply separate out the common alphabet. in CORPORATION , o is 3 times and r is two times, putting together we have,OOORRCPATIN . now the total unique alphabets are 7 and thus the answer is 7%=7*6*5*4*3*2*1= 50400 . the common alphabets dun need any permutation. |

Venumadhav said: (Dec 28, 2010) | |

Actually if you consider ooo as o and rr as are the there are 8 distinct alphabets. C O R P A T I N. |

Kumaran said: (Jan 6, 2011) | |

How the 7 is came instead of 6 (CRPRTN). |

Raghav said: (Jan 26, 2011) | |

What is process when same letter is come in like as this question? Thanks. |

Suja said: (Feb 12, 2011) | |

We consider all vowels as one letter i.e. OOAIO is a letter + rest of the 6 letters = 1+6 =7. |

Divya said: (Mar 10, 2011) | |

Good explanation. Thank you. |

Vinod said: (Jul 23, 2011) | |

Why divide 7! by 2! ? because you can treat the problem as a permutation with subgroups of identical items. the general formula is nPn1,n2,n3... equals n! divided by n1!n2!n3!... In this problem you have n = 7 letters (6 plus the vowel group). two letters are the same so n1 = 2. the rest are unique so the 5 other subgroups = 1. so you 7! divided by 2!1!1!1!1!1! . the answer as given simply didn't write out the 1! terms. |

Kanakam Vinay Kumar said: (Aug 11, 2011) | |

I understood Very Clearly. |

Smita said: (Mar 16, 2012) | |

How many ways are there to divide 5 toys among 3 children so that each child gets at least 1 toy? |

Ankit said: (Apr 23, 2012) | |

(5!/3!) and (7!/2!) is due to the reason that in the vowels O comes 3 times therefore it has to be in the P&C part too. same case with consonants |

Gaurisha said: (Aug 22, 2012) | |

Thanks vinod. |

Abhishek said: (Sep 17, 2012) | |

Why we divide the vowel from all letters?and 1 more why we also calculate the vowel group also because we consider vowel as 1 and we had adjusted as a 7!. Then why again? |

Shashank said: (Mar 19, 2013) | |

Now, let me explain in brief for those who don't find it easy. Given word is 'CORPORATION' Total Alphabets = 11. vowels = 5 i:e(O,O,A,I,O) Consonants = 11-5=6 i:e(C,R,P,R,T,N) Now,treating vowels as 1 alphabet as asked in problm we hv=6+1=7 Also alphabet R comes twice . Thus from "IMPORTANT FORMULAS" NO.4 For (C,R,P,T,N+(O O A I O)) p1=C(1) p2=R(2) p3=P(1) p4=T(1) p5=N(1) p6=(O O A I O)(1) (p1+p2+p3+p4+p5+p6)= n i:e 1+2+1+1+1+1=7. Thus, n!/(p1!).(p2)!.....(pr!)= 7!/1!.2!.1!.1!.1!.1!=7!/2!=2520. Similarly, vowels (O,O,A,I,O) can be arranged in p1=O(3) p2=A(1)p3=I(1) p1+p2+p3 i:e= 3+1+1=5 5!/3!.1!.1!=5!/3!= 20. Now the word 'CORPORATION' can be arranged in= 2520*20=50400 ways. |

Math Wiz said: (Apr 21, 2013) | |

Every repeated vowels and consonants, you must divide them with the corresponding numbers if it. Like the one up there, O is repeated thrice (3 times) so you must divide them with 3!. Same goes to R which is repeated twice. That's they must be divided with 2!. After that you can just multiply both of permutation, 7! and 5! which were already divided by their repeated letters. |

Sirius said: (Jul 9, 2013) | |

There is some flaw on the questions, it should be stated "all vowels". If not, it can mean 3 vowel together and the other 2 vowel group together but this 2 group can separate differently. Example O O_ _ _ A I O _ _ _ . |

Baz said: (Sep 25, 2013) | |

Good day! How did you get 2520 and 20 respectively because if u divide 7/2 and 5/3. You will get this 3.5 and 1.7. |

Belimgoe Emmanuel said: (Mar 27, 2014) | |

How come we had 5/3 = 20ways? |

Arvind said: (May 7, 2014) | |

For @Baz and @Emmanuel .Its not 5/3 its 5!/3! (factorial ways). The formula for the division of this is give as a result in Important Formulas prescribed in this site (Formula 4). But rather than the formula it is more of Division Logic. This Division logic comes more often in chapters of Prob and statistics. " Logic - for the vowel letters OOAIO, consider only OOO. In 3 spaces only 1 combination can be made, so in 5 spaces (OOO_ _ )how many combination can be made? hence 5!/3!. Division is cummulative subtraction ( therefore in 3! ways only 1 way is the right answer and other all 3! - 1 are wrong answers) |

Kiran said: (Jun 24, 2014) | |

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always not together. |

Ramya said: (Jul 7, 2014) | |

How you get 2520? |

Gerald said: (Jul 31, 2014) | |

Please I'll like to know how you get 2520 and 20? |

Sindhuja said: (Aug 29, 2014) | |

If we have more than one consonants or vowels being repeated again. Then how does the calculation will be? Eg : Appetite. |

Hema said: (Sep 11, 2014) | |

How will be7!/2! = 2520? |

Raj said: (Oct 7, 2014) | |

Hi friends actually that is a one formula npr formula is used. |

Preeti said: (Dec 6, 2014) | |

Which formula is used in 7!/2!? |

Kirti said: (Mar 2, 2015) | |

7!/2! = 2520. How is possible? |

Ram Mohan said: (Apr 21, 2015) | |

Hi. In the above {CRPRTN (OOAIO)} (OOAIO can be treated as one letter). Out of those 7 letters can be arranged in 7! ways. But there are 2 R's is there. They can exchange their places. That's why we are doing 7!/2!. 7! = 7*6*5*4*3*2*1 = 5040/2! = 2520. In the vowels similarly 3 O's can exchange their places. 5!/3! = 5*4*3*2*1/3*2*1 = 20 ways. = 2520*20 = 50400. |

Rajmachawal said: (Jul 19, 2015) | |

Shouldn't there be a multiplication by 2fac in the end because there can be vowels first and then consonants or the consonants first and vowels second? |

U. Sathish Reddy said: (Aug 31, 2015) | |

You can do either of the way either consonants first or the vowels first. Just separate the Word CORPORATION along with the vowels i.e CRPRTN: These are 6 consonants, as are is repeated two times you need to divide it by 2! Hence, 6+1 = 7!/2! = 5040/2 = 2520. 1 was from the set of vowels. (OOAIO) - These are 5 vowels which is represented as 1 set. Now, from the Vowels we get. 5!/3! = 120/6 = 20. i.e 2520*20 = 50400. |

Arun Kumar said: (Oct 4, 2015) | |

Which explanation is correct? |

Ambrose said: (Oct 18, 2015) | |

Given a set [ABCDEFGHIJKLMNOPQW]. In how many ways can we form a non repeated sub-set of 4-letter with letters (AB) being constant in every sub-set. Example [ABCE], [ABDC]. |

Vishnupriya said: (Jan 29, 2016) | |

Find the number of ways in which an arrangement of four letter can be made from the letters of PRECIPITATION? Please help me to solve this problem. |

Aarti Gupta said: (Feb 21, 2016) | |

What is the difference between permutation and combination? I seriously can't understand the logic. |

Aarti Gupta said: (Mar 2, 2016) | |

Is there someone who can help me out to understand the difference between both of them. |

Bob said: (Mar 6, 2016) | |

Where in question 3 does it indicate that only one vowel and only one consonant from the alphabet are permitted? Since this question appears to duplicate the wording of question 2, why isn't it solved in a like manner ? What am I missing here ? |

Abhi said: (Aug 12, 2016) | |

How come 7!/2! Is 2520? 7!/2! = (7*8) / (2*1) = 32 isn't? Then how 2520? |

Abhi said: (Aug 21, 2016) | |

@Aarthi, The combination is selection only whereas permutation is selection + arrangement. For example: A committee should be formed with 2 persons from A, B, C, therefore, the should be {AB, BC, CA} only not {AB, BC, CA, BA, BC, CA} because AB and BA both are the same combination. Next, If the question is like committee should be formed with one president and one vice president the answer is {AB, BC, CA, BA, BC, CA} because, in the combination AB, A can be president and B can be vice president and vice versa. So both are different. |

Venkata Dinesh said: (Sep 6, 2016) | |

The give word consistent of 5 vowels and remaining 6 letters + 1 group of vowels is 7!. So to get answer we must do 5! * 7!. Is this correct? I am getting wrong answer. Please Help me. |

Keerthana said: (Sep 28, 2016) | |

But here, R is repeated 2times, in vowels O is repeated 3 times. So we have to consider R & O. So, 7!/2! & 5!/3! is taken. |

Chandu said: (Oct 7, 2016) | |

I understood. Thank you all. |

Sofia said: (Nov 24, 2016) | |

Where did '7' came from? |

Rahul Jha said: (Dec 17, 2016) | |

@Sofia. We have to consider all the vowels as 1 vowel + rest 6 consonants = 7. |

Neethu said: (Dec 29, 2016) | |

CORPORATION Vowels Come Together. Consonants CRPRTN ( R repeat twice). Vowels OOAIO ( O repeat thrice ) We take 7!/2! * 5!/3!. Note: 7! means the number of consonants + the vowels taken as "one". that means 6 + 1 ( CRPRTN + ( OOAIO)). 7!/2! means R repeated twice, so we take 2!. 5!/3! means O repeated thrice, so we take 3! too. |

Maurice said: (Jan 8, 2017) | |

Please one should help me out with this complicating question: (a) How many four and five digits numbers can be formed using 2, 3, 4, 5, 6? (b) How many will be greater than 5,000? (c) How many will be odd numbers? (d) How many will be even numbers? PLEASE show workings thank you. |

Gbenga Felix said: (Feb 24, 2017) | |

A school principal and his wife, as well as three other tutors are to be seated in a row so that the principal and his wife are next to each other. Find the total number of ways this can be done. Please give me the answer. |

Naveen said: (Mar 12, 2017) | |

@Gbenga. Total there are 5members while 2are always together so considering 2members as -unit arrangement can be done in 4! Ways and they themselves i.e the principle and wife can be arranged in 2! ways so total no of ways will be 4!*2! = 48ways. |

Shobhit said: (Mar 13, 2017) | |

@Maurice. a. 5 digit number can be arranged in 5! Ways 4 digit number can be arranged in 5c4 * 4! Ways. |

Madhu said: (May 7, 2017) | |

Why the answers 2520 and 20 are multiplied why not they added? Please give me the answer. |

Shibu said: (Jul 26, 2017) | |

Hi @Madhu. From the formula, we get only the multiplication so we are doing the same. |

Raghvendra Patel said: (Sep 19, 2017) | |

Should it not be 5!/3! * 6!/2!* 2!. As first arrange vowel and then consonant separately and then arrange these two packets. |

Lokesh said: (Oct 16, 2017) | |

Yes, there is a logic. The word corporation 'r' can be repeated two times so it is 2! And in vowels 'o' can be repeated three times. So it is 3! |

Aswathy said: (Sep 17, 2018) | |

Thank you for the given explanation. |

Pavi said: (Sep 18, 2018) | |

I am not getting. Can anyone help me to solve this? |

Veera Manikantha said: (Sep 26, 2018) | |

In CORPORATION their vowels have to be together, so it becomes CRPRTN (OOAIO). In this, a number of ways is 7!÷2! since R is repeated 2 times and in the vowels, there are 5!÷3! A number of ways. So total number of ways is (7!÷2!) * (5!÷3!)=50400. |

Mayuresh Hedau said: (Jul 8, 2019) | |

Firstly there are 11 words(5 vowels and 6 non vowels). Consider 5 vowels =1 and remaining words are 6. Add them (6+1=7). In those 7 words count the Repeating words and divide to 7 (likewise 7fact÷2fct) and, In those 5 vowels count the repeating words to and divide to 5 (5fact÷3 fact). Multiply both of them : (7fact÷2fact) * (5fact÷3fact). = 50400. |

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