### Discussion :: Permutation and Combination - General Questions (Q.No.2)

Shoba said: (Sep 4, 2010) | |

Other than vowels there are only 4 letter then how it s possible to get 5!. |

Sai said: (Sep 10, 2010) | |

HI SHOBA,. 4 consonants + set of vowels (i. E. , L+N+D+G+ (EAI) ). We should arrange all these 5. So we get 5!. I think you understood. |

Sachin Kumar said: (Sep 29, 2010) | |

Please make me understand this answer as did not get. |

Sri said: (Oct 10, 2010) | |

As vowels are together take (EAI) as single letter i.e. , total no of letters are 5 (L, N, D, G, {EAI}). No of ways can arrange these 5 letters are 5! ways. Now we arranged 5 letters (L, N, D, G, {EAI}). Next we have to arrange E, A, I (they may be EAI/EIA/AEI/AIE/IAE/IEA). All these combinations imply that vowels are together. So we have to multiply 5! and 3!. |

Subbu said: (Jan 6, 2011) | |

7!=5040 |

Madhusudan said: (Mar 14, 2011) | |

Could you kindly let me know what is ( ! ).Howe 5! = 120 ? |

Sundar said: (Mar 14, 2011) | |

@Madhusudan 5! = 5 Factorial = 1 x 2 x 3 x 4 x 5 = 120 |

Laxmikanth said: (Mar 26, 2011) | |

When should we take the one or more letters as a single unit and why? |

Moaned said: (Apr 8, 2011) | |

I am not getting |

Jessie said: (Apr 10, 2011) | |

7 letter word = LEADING CONDITION = VOWELS TO BE TOGETHER, HENCE (EAI) TO FOR A WORD SO NO. OF WORDS = L,(EAI),D,N,G = 5 permuation to arrange 5 letters = nPr= n!/(n-r)!=5!/0!=5! 0! is assumed to be 1!) EAI can be arranged among each other in = nPr = 3!/(3-3)= 3! hence 5! x 3! = 120 x 6 = 720 |

Mayur said: (Apr 15, 2011) | |

How it came like nPr formula and how you have solve it? please let me know. |

Bharti said: (Jul 11, 2011) | |

Thanks a lot. I had confusion before your explanations. Thanks a lot. |

Pavan@9966606261 said: (Aug 5, 2011) | |

@mayur. You just look when ever you open the new exercise there will be availability of basic formulas. If you go through them half of the task would be finished easily. Have a good day buddy. |

Siva said: (Oct 16, 2011) | |

Permutations and combinations always make me to confuse much. How to decide based on the descriptive aptitude question ? |

Shiva said: (Oct 18, 2011) | |

Why we have taken 5 (4 + 1) ? |

Santosh Kumar Pradhan said: (Oct 27, 2011) | |

Total member 7 out of which 5 consonant and 3 vowels take 3 vowls as a 1 hen number of consonant will arrange 5! ways and these 3 vowels will arrange 3! ways though,5!*3!=720 |

Bhavik said: (Nov 4, 2011) | |

How to read these nPr ? |

Sagar Choudhary said: (Nov 6, 2011) | |

What is this 5! and 3! ? |

A.Vamsi Krishna said: (Jan 30, 2012) | |

"!" this implies factorial that means a number is multiplied like for example take number 5 then its factorial will be taken as 5*4*3*2*1 and this is equal to 120. |

Ncs said: (Feb 2, 2012) | |

Why not 5! + 3! ? |

Raj said: (May 21, 2012) | |

Friends, How do you say this question is permutation. |

Mahanthesh said: (Jul 30, 2012) | |

Hi guys. As per the question, it s mentioned that all vowels should be together, but it has not been mentioned that it should be EAI. According to me these 3 vowels can be arranged in 3*2=6 ways. And the remaining letters LDNG can be arranged in 4*3*2*1= 24 ways. Can anyone explain me about this please ? |

Sandeep said: (Sep 7, 2012) | |

can any one explain ? y 5!*3! & y not like this 5!+3! |

Srk said: (Oct 9, 2012) | |

@Mahantesh. 1. There is given a word LEADING in this LDNG (consonents) , EAI (vowels). 2. They asked here vowels always come together and so we should have to take LDNG (EAI). 3. We can take as (4+1) ! i.e. here we have to take vowels as together as 1. So we have a chance of 5!. 4. But with in vowels we have many arrangements i.e 3! 5. Finally 5!*3! |

Shafi said: (Oct 11, 2012) | |

@Mahantesh, You are correct 3! and 4!, because you spitted vowels and consonants, But to combine them vowels+consonants. i.e. (4 consonants + 1 set of vowels) = 5! And (3 vowels {1 set}) = 3! So 5!*3! = 5x4x3x2x1x3x2x1=720. |

Adhityasena said: (Feb 9, 2013) | |

I get the answer to be 2*720. Because, since the vowels must come together in the word LEADING, which actually has 7 letters, E and A can be taken as one unit, so now we have L, EA, D, I, N, G to be arranged and which can be done in 6! ways. But among E and A there are 2 arrangements namely EA and AE, So the final answer is 2*720. Am I right ! |

Chetas said: (Jul 6, 2013) | |

@Adhityasena. You have not considered a vowel 'I'. 'EAI' is to be taken as one unit. |

Sanjay(9776387850) said: (Aug 29, 2013) | |

Hi Guys, at first I tell you how can you know is this permutation or combination. Permutation means arrange(row/column) and Combination means selection(group). i.e. Number and Word are perm. Playing 11 and committee are combi. This is a word so this is perm. You should know the formula that m different objects are alike and n different object are alike if we arrange all the m+n objects such that n objects are always together=(m+1)!*n! Here n = Vowel = 3 and m = Con. = 4. So = (4+1)!*3! = 5!*3!. = 120*6 = 720. |

Chithra said: (Nov 5, 2013) | |

As we known very well that * is consider to be AND, + is consider to be OR. Consider the 1st example 7 men and 6 women problem we taken as (7c3*6c2) + (7c4*6c2) + (7c5). We can also said this as (7c3 AND 6c2) OR (7c4 AND 6c2) OR (7c5). CONDITION -> 5 people need to select. We should take 3 men from 7 men and 2 women from 6 women or other option is to take 4 men from 7 men and 1 women from 6 women. That's why we are using * at the place of AND, + at the place of OR. Similarly, we need to consider both the consonant AND vowel not consonant OR vowel. So we use 5!*3!. |

Taku Mambo Bellnuisemarbel said: (Nov 29, 2013) | |

Please help me with this; whenever I hear of probability that a variable is being selected what should I think of? |

Amna Fida said: (Dec 3, 2013) | |

How we know that factorial will use here? |

Amnafida said: (Dec 3, 2013) | |

Describe that how we know about here permutation is use? |

Jhansi Sri said: (Feb 7, 2014) | |

Please help me quickly why we take 5!*3!, Why we can't take 5!+3!. |

Pema said: (Aug 3, 2014) | |

When it comes the question for arrangements, then it is a Permutation Or you all can remember it as keyword "PA" P=permutation and A=arrangement. Likewise, for combination, it is all for selection purpose, remember keyword as "CS" c=combination,s=selection. Then apply formula for each. Easy. |

Baidyanath Jena said: (Nov 7, 2014) | |

When it comes to persons it should be combination. |

Samson said: (Nov 17, 2014) | |

God bless you all for your contribution especially you @Jessie for using the formula to break it down well. |

Ranjeet said: (Nov 19, 2014) | |

Well I am confused. Somewhere n! is done whereas somewhere (n-1)! is used. Can someone explain about it? |

Sagar said: (Dec 26, 2014) | |

Hi friends. We know that formula n!=n (n-1) (n-2).....3.2.1. Suppose there n way to choose first element (since there are n elements). After that there are n-1 ways to choose second element because already we choose one element from n elements that's why we are assuming this way. Similarly n-2 ways to chose the third element..etc it's going like this. n!=n (n-1). n!=n (n-1) (n-2) if n>2 or equals 2. n!=n (n-1) (n-2) (n-3) if n>3 or equals 3. Hope you understood. |

Shantha said: (Apr 30, 2015) | |

Then how we won't take E+A+I+(LDNG) = 4. |

Tom said: (May 2, 2015) | |

In what situations we can permutation or combination? |

Riya said: (May 8, 2015) | |

@Tom and @Shantha. Whenever there is a reference to some arrangement it is permutation. Whenever there is a reference to some selection it is combination. The given question requires us to find the number of ways in which the word LEADING can be arranged with the condition that the vowels (EAI) always be together. Thus we need to apply the concept of permutation. Here, since the vowels EAI must always be together we consider it as a single word (EAI). Thus, LDNG (EAI) make a 5 letter word. It can be arranged in 5p5 ways = 5!ways. Now, since EAI can arrange itself in 3p3 or 3! ways, the word LDNG (EAI). Can thus be arranged or PERMUTED in 3!*5! ways = 720 ways. |

Spurthy said: (Jun 24, 2015) | |

All the consonants can also be written as a unit? |

Ajay said: (Aug 17, 2015) | |

In some cases unit's place, tens place's and hundred's place are used what its mean? |

Shrinivas said: (Aug 25, 2015) | |

Hi if two vowels are repeated in same word then will you take it as same or as different? |

Vanmathi said: (Sep 1, 2015) | |

In how many ways LEADING be arranged such a way that atleast two vowels always together? |

Soumya Sengupta said: (Sep 18, 2015) | |

If there is another vowel 'o' what should be done like in 'outstanding' here a, i, o, u are vowels do we have to consider (aiou) = 1 letter for calculation? |

Palak said: (Sep 18, 2015) | |

In this question won't the arrangement of non vowels matter and why? |

Raja said: (Sep 19, 2015) | |

Hai @Soumya. S its 1 letter only. Hope you understand. |

Yonatan said: (Sep 24, 2015) | |

I can't understand the answer. |

Mukesh said: (Dec 18, 2015) | |

Please anyone help me why repetition is not considered in the current problem? |

Eswaru said: (Feb 3, 2016) | |

Hey dude we have to form 7 letter words from LEADING that means we need to use all the letters at a time. So no repetition are allowed. |

Peter said: (Feb 28, 2016) | |

If at least two vowels always together then what will be the answer for LEADING ? |

Siva said: (Mar 16, 2016) | |

Other than vowels there are only 4 letter then how it s possible to get 5!. |

Brian said: (Jun 2, 2016) | |

Correct me if I'm wrong but, There are 7 letters: L E A D I N G 3 vowels and 5 non-vowels. I got the part where we need to permutate 3 and 5, resulting with 3!*5!. But aren't there also several other positions that these vowels could be positioned? For Example: AIELDNG is one factor, another factor is LAIEDNG, another factor is LDAIENG. As you can see, the positions of the 3 vowels with each other are the same, and the order of non-vowels in the word is also the same, but I only changed the position of the starting point for the vowel permutations, starting from the first position, to the second, to the third. So, I believe that the answer should be 3!*5!*5. Since there are 5 different ways you could represent the same order of vowels in different positions with the same order of non-vowels. Please do correct me if I'm wrong or if I misunderstood the question |

Technothelon said: (Jun 9, 2016) | |

@Brian. There aren't such kind of ways. Because arrangement is not taken into consideration when we do combinations. |

Kirupa Rani D said: (Jun 29, 2016) | |

Can you give solution for this problem? How many words can be formed from the letters of the word 'PACKET', so that the vowels are never together? |

Jomson Joy said: (Aug 3, 2016) | |

In PACKET there are 2 vowels. Vowels came 2gether means 4! * 2! = 240. Total words formed =6! (because of total letters) = 720. Therefore 720 - 240 = 480. |

Gopika S S said: (Aug 3, 2016) | |

A number of permutations of "n" different things are taken "m" specified things always come together is m!* (n-m+1) ! Here the three vowels always come together. So here m is 3. A total number of letters is 7. Substitute we get. 3!* (7 - 3 + 1)! = 3! * 5! = 6 * 120 =720. Hope you got it. |

Prajwal said: (Aug 17, 2016) | |

Does it have an another method to find the solution? |

Rakesh said: (Sep 16, 2016) | |

In leading 5! * 3! ways we can fill all vowels come together. |

Otieno said: (Nov 30, 2016) | |

Where has 5! come from? Please explain me. |

Xyz said: (Dec 20, 2016) | |

How many letters can be formed from COMBINATION if vowels are kept together? Please give the solution. |

Sowmya said: (Dec 23, 2016) | |

@Xyz. C,M,B,N,T,N,(O,I,A,I,O)=> C,M,B,T,N,(O,I,A)=> 6! ways. Vowels alone can be rearranged themselves in 3! ways. So 6! * 3! = 2160. Hope this is right. |

Navan said: (Feb 2, 2017) | |

I have one question. How many ways 11players selected from 15 players? Please give me the answer. |

Akshay said: (Mar 31, 2017) | |

@Navan. 15c11 = 15!&div(15-11)! * 11! |

Avishek Kadel said: (Apr 21, 2017) | |

In how many ways the letters of word ELEMENT can be arranged so that vowels are always together? Can anyone solve this? |

Ali said: (Apr 26, 2017) | |

@Avishek No. of vowels is 3Es and the can only be together in this pattern {EEE}. There is only 1 way of choosing so 1! = 1 * 1 = 1. We are to consider 3Es as a single alphabet because they are together. So we have L, M, N, T,{EEE} ie 5 letters now in all. Now of way of choosing 5 letters = 5! (5 * 4 * 3 * 2 * 1) = 120. Finally, we multiply 120 * 1 = 120 ie. the number of ways of arranging the letters so that the vowels are always together. |

Krishna said: (May 23, 2017) | |

LEADING- vowels together in total 5 positions at L, E, A, D, and I. Ex. For the first position - (EAI) - (remaining 4 lettersLNGD) - 4! * 3! For total 5 positions - 5*3!*4! -720. |

Abhishek said: (Jul 18, 2017) | |

I think it should be 3!*4! |

Tanmay said: (Aug 14, 2017) | |

In how many different ways can the letters of the word software be arranged in such a way that the vowels always come together? |

Jon said: (Aug 17, 2017) | |

@Tanmay. SFTWR (OAE) 5 +(1) ! = 6! OAE can be arranged in 3 ways 3! 6! = 6*5*4*3*2*1 = 720 ways 3! = 6 ways 720*6 = 4320 ways! |

Dhana said: (Oct 9, 2017) | |

Word : leading condition : vowels together eai-3! ldng&eai-5! (ldng)&(eai)-2! Whether the last condition is valid? Give explanation. |

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