# Aptitude - Permutation and Combination - Discussion

### Discussion :: Permutation and Combination - General Questions (Q.No.5)

5.

In how many ways can the letters of the word 'LEADER' be arranged?

 [A]. 72 [B]. 144 [C]. 360 [D]. 720 [E]. None of these

Explanation:

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

 Required number of ways = 6! = 360. (1!)(2!)(1!)(1!)(1!)

Video Explanation: https://youtu.be/2_2QukHfkYA

 Shweta said: (Jun 18, 2010) Why the answer 720 is incorrect?

 Abhishek Krishna said: (Sep 4, 2010) Hi Madhu and Shweta, the answer is right if see I tell how the letter E is repeated two times so it has to be divided and rest of letters are only one times occurred, so dividing from 1 did not make any thing.

 Neeraj said: (Sep 8, 2010) Hmm Krishna you are right. I got the answer by your way.

 Sunil said: (Nov 16, 2010) Explain Please

 Sam said: (Feb 13, 2011) Leader has 6 letters so it can be arranged in 6! ways but letter e is repeated twice so ans is 6!/2!=360.

 Ritu said: (Apr 9, 2011) 6*5*4*3*2*1=720 720/2=360

 Umakant said: (May 12, 2011) Why is wrong 5x4x3x2x1?

 Umakant said: (May 12, 2011) In this Method 6*5*4*3*2*1=720 720/2=360 Why divide by 2 Please explain it

 Praveen said: (May 31, 2011) We can't divide by 2 may be what not 5!?

 Sonu said: (Jul 14, 2011) Ritu explain it in a very gooo way thanks a lot ritu.

 Vinny said: (Jul 18, 2011) LEADER contains 6 letters. So 2 are similar letters. Therefore 6!/2!=360.

 Poornima.R said: (Aug 26, 2011) Why we want to divide this 720 by 2?

 Achutharaj said: (Sep 28, 2011) Hi poornima, E is repeated in two times so we divided leader -l,e,a,d,r ( e is repeted 2 times so considered single time) 6!=6x5x4x3x2x1/2x1=360

 Tezan said: (Sep 30, 2011) Why we take 6!/2! Whay we have not take vowel 3 and consonent 3. like (3+1)4 4!= 4*3*2*1= 24 3!/2!=3 24*3= 72

 Lalit said: (Jan 23, 2012) Helo frndz in this word LEADER thre r 6 leters in which 2 leters are repeted thats why we divide it by 2! and the ans is 6!/2!

 Tanaya said: (Feb 1, 2012) I have query that all these problems belong to combination and not permutation section?

 Abhishek said: (Sep 17, 2012) Why not 5! * 2!? because E is repeated so now consider 5 letters so 5! and E is repeated twice so 2! ??

 Sujit said: (Nov 27, 2012) @Tanaya: both permutation and combination are used in this section. In some problems we arrange the letters means they are of type permutation..and in some cases like selection group we used combination :-). @Abhishek: let me explain you with the example Suppose your name is : AAB which contain A 2 times (in above question E repeate 2 times so i consider name with one character repeat twice) AAB can be arranged in AAB ( 1) ABA (2) BAA (3) Means 3 ways.. This ans can be calculate directly using formula Formula: (number of characters)! ------------------------ (number of repeat characters) ! In AAB there are 3 characters and one character repeat twice So above formula become 3! 3*2*1 -- = ------ = 3 which we got above by doing manually 2! 2*1 similarly in above question LEADER contains 6 character and E repeat twice so using formula 6! --- = 360. 2! Hope you got this :-).

 Jonathan said: (Dec 6, 2012) LEADER One way of arranging this is "EADERL" (The first 'E' and the second 'E') If I want to put the second 'E' in place of the first 'E' to make another arrangement, I would get "EADERL" which is the same as the previous one. POINT: we have shown that eventhough I interchange the first 'E' and the second 'E', we came up with the same arrangement. MEANING... if there are TWO same letters, there will be DUPLICATION... That is the reason why we DIVIDE by TWO! It's not 5! because still, there are six letters we are arranging! Hope this insight can help you!:)

 Sundar said: (May 3, 2013) In how many different ways can the letters of the word 'ORANGE' be arranged so that the three vowels never come together ? Answer : The total number of ways of arranging ORANGE = 6! The total number of ways o,a and e can be arranged = 3! The total number of groups when the vowels are one group and the rest are individuals= 4! = 6! - 4! x 3! = 576 In how many different ways can the letters of the word 'EXTRA' be arranged so that the two vowels never come together ? Answer: The total number of ways of arranging EXTRA = 5! The total number of ways e and a can be arranged = 2! The total number of groups when the vowels are one group and the rest are individuals XTR(EA) = 4! = 5! - 4! x 2! = 72

 Fritz said: (Jul 16, 2013) This problem is combination or permutation ?

 Vikram said: (Aug 31, 2013) 360 is the correct answer because "C" is repeated 6!/2! = 6*5*4*3 = 360.

 Pavithra said: (Dec 24, 2013) Why we should take like 6*5*4*3?

 Ashwini said: (Jan 21, 2014) Hi @Pavithra the word LEADER has repeated letter 'E' twice so 6!/2! = 6*5*4*3 = 360.

 Sankari said: (Jan 23, 2014) In how many ways ELECTION can be arranged so that the vowels occupy the odd places.

 Thokie said: (Mar 14, 2014) There are 9 switches in a fuse box, how many different arrangements are there?

 Mancy said: (Apr 25, 2014) Hi @Thokie, Q: There are 9 switches in a fuse box, how many different arrangements are there? Ans: 2 2 2 2 2 2 2 2 2. Each switch has two possible, on or off placing a 2 in each of the 9 position. So, we have 2 raise to the power 9 = 512. So it means there are 512 different arrangements can be done. Hope this'll help.

 Swapnil said: (May 29, 2014) In this letter e repeated twice. So that in permutation swapping of e letter. Does not form new word and so they are counted. Once at every case so divided by 2!.

 Yusuf said: (Jun 12, 2014) Why not sometime division and sometime by considering vowel?

 Harish said: (Jul 2, 2014) In how many different way can the letters of the word CREAM arranged.

 Pri said: (Aug 8, 2014) Actually 6 letter so we got 6*5*4*3*2*1= 720 but e 2 time repeated so we divided 720/2=360.

 Gopal said: (Aug 23, 2014) I have a similar doubt, I have 2 each of 24 different colors, how many 12 color combinations are possible?

 Veenita said: (Nov 24, 2014) First the word letter it can come in 5 different ways and the word letter have 6 words.

 Anand Khakal said: (Dec 11, 2014) Please explain following question I am not understand what is mean by 6! = 360? (1!) (2!) (1!) (1!) (1!). !- what is this?

 Suleman said: (Mar 14, 2015) It is a permutation question, in which some letters are repeated so it can be solved as, ans = 6!/2! =720/2 = 360.

 Pavan Kalyan said: (Apr 3, 2015) Yes friends answer 360 is correct because E is repeated 2 times.

 George said: (Jun 24, 2015) Lets look at it this way: In LEADER, there are 2 Es and 4 Unique alphabets (LADR). Now selection of 2 Es in any of the 6 positions is a combination problem. Hence E can be selected in 6C2 ways. Which is 6x5/2! = 15. For each of these 15 choices of E, there are 4 remaining slots for 4 unique alphabets. Hence this is a permutation problem. The 4 unique alphabets can be arranged in 4! ways i.e. 4 x 3 x 2 x 1= 24. Therefore total ways of arranging is 15x24 = 360.

 Rohan said: (Jul 21, 2015) It is asking how many way not how many different ways. So, answer should be 6! = 720.

 Kuldip said: (Aug 12, 2015) Can anybody explain me when permutation & when combination used?

 Mukund said: (Aug 19, 2015) Whether we need to divide by 2 or 2!. As in this, it doesn't matter but if "e" was repeated thrice, then do we need to divide numerator by 3 or 3!

 Kelvin Esekon said: (Oct 5, 2015) The word as 6 letters, the letter e is repeated twice, so: npr = 6!/2! = 360.

 Paulvannan said: (Mar 5, 2016) Hi factorial value is 1*2*36*4*5*6* = 720. So 720/2 = 360.

 Jonse said: (Mar 27, 2016) 6!/2! = 6*5*4*3*2*1/2. = 720/2. = 360.

 Savitri said: (Apr 6, 2016) @Sundar, Can you please explain this, how it is 72? = 5! - 4! x 2!. = 72.

 Savitri said: (Apr 7, 2016) (1!)(2!)(1!)(1!)(1!) How it is? Explain me.

 Divya said: (May 15, 2016) @Savitri. It is; L = 1. E = 2. A = 1. D = 1. R = 1.

 Dipankar Rabha said: (May 22, 2016) How did 360 come? What is (1!)? What is the meant of this symbol? And can anyone provide the easy method for this type of problem?

 Sandeep said: (Jun 30, 2016) I have a query. LEADER= (LADR) +) (EE) = 4 + 1 = 5! = 120, and divided it by 2C2!, why can't be the answer 120. The same way for CORPORATION problem is solved. All vowels are a set & consonants are of another set and as the vowels are scrambled it was divided by vowels combinations.

 Manoj D Desai said: (Jul 14, 2016) Please See the question once carefully, it's just asked for number of ways so word repetition is not considered so the correct answer is 720 (6!) , If the question is asked like " How many DIFFERENT ways word Leader can be arranged" then it would have been 6!/2 for E-letter repeating 2 times.

 Sanjay P said: (Jul 27, 2016) I have to plan audit in such a way that 24 activities are to be assessed in 6 different laboratories in a year with 9 assessors. Two assessors are required for on activity.

 Kumar said: (Sep 1, 2016) LEADER has 6 words totally. But letter E repeatedly two times then, to get a required solution we use 6! (total words) /2! (repeated word). So, the answer is 360.

 Nanda said: (Oct 16, 2016) Why shouldn't the vowels and consonants be grouped differently ? like LDR (EEA) which makes it 3 + 1 = 4! And 3!/2! since E has repeated twice like in the previous problems.

 Anil said: (Nov 29, 2016) Total letters-6! And repeated letters 2 so-2! Then 6!/2! = 360 right answer.

 Vikram said: (Dec 9, 2016) Take the example of the word EYE: It can be arranged in 3 ways only , namely EYE , EEY, YEE as the letter E is repeated twice. So the answer is not 3! but 3!/2!. Applying the same for LEADER, we have the answer 6!/2!.

 Shivaraj said: (Dec 18, 2016) LEADER It contains 6 letters i.e=6! Now take repeated letters i.e E(2 times), so it can be divide by 6! i.e LEADER =6!/2!. = 6 * 5 * 4 * 3 * 2 * 1/2 * 1 = 360 So,360 is correct answer.

 Shivaraj Kavalaga said: (Dec 18, 2016) @Vikram. EYE It contains 3 letters i.e 3! Now take reapeted letters i.e E(2times repeated)=2!, EYE = 3!/2!, = 3 * 2 * 1/2 * 1, = 3.

 Shivaraj Kavalaga said: (Dec 18, 2016) @Savitri and @Divya. Don't go that method because you'll get big confusion in solving big problems.

 Isha said: (Jan 7, 2017) It is 720.

 Khan said: (Jan 20, 2017) In how many ways the letter of the word SAMPLE be arranged if all the latter are taken? Please slove with formula.

 Prakash said: (Feb 2, 2017) Why we use permutation formula in this example? Is there any way to solve this?

 Gowtham said: (Mar 1, 2017) @Khan. The word SAMPLE has 6 letters so we apply the formula in 6!= 720 ways can we arranged. Am I Right?

 Shekhar said: (Mar 6, 2017) In how many ways can the letters of the word 'DIRECTOR', be arranged so that the vowels are never together? Please solve this.

 Ezaz Ahmed said: (Mar 31, 2017) Thanks for giving the explanation of the answer.

 Vamshi said: (Jun 26, 2017) They didn't ask without repetition? Then why you divided by 2?

 Parthiban said: (Jul 4, 2017) LEADER : totally 6 words so= 6!. L1, E2, A1, D1, R1 so= 1!*2!*1!*1!*1!. 6!=6*5*4*3*2*1=720. 1!*2!*1!*1!*1!= (1*1) (2*1) (1*1) (1*1) (1*1)=2. Total words=720/2=360.

 Sanjib said: (Jul 14, 2017) LEADER : totally 6 letters. so= 6!. And vowel 3 letters (2->E and 1->A) so we can solve in this way : 6! /2!=6*5*4*3*2! /2!=360 (ans).

 Lavanya said: (Aug 15, 2017) Without mentioning why did they do with repetition. Why can't we do without repetition?

 Shwetha said: (Aug 19, 2017) Hello, @All. In this question it means like how many ways (it means without repetition of letters) the "LEADER" letter can be arranged it supposed to be 6!=720. If the question is like how many different ways (with repetition of letters) can letter "LEADER" arranged then 6!/2! (2! is because of E letter repetition) = 360. Hence the solution for this problem should be 720 as per my knowledge.

 Rdg said: (Nov 13, 2017) Sometimes vowel sometimes that way nothing could be understood please explain in sequence. And introduce the differences of combina and permu. Try to teach from a novice perspective.

 Manish said: (Nov 23, 2017) 7 letters out of which 2 letters are repeated. Therefore 7!/2!

 Manish said: (Nov 23, 2017) If vowels point is given then vowel to be treated as one-word & if that one word contains repeat letters they must be divided. For eg- eeo. 3!/2!.

 Asam said: (Jul 17, 2018) Please explain this.

 Anandavalli said: (Sep 4, 2018) But E and A are vowels so we have to take it as 1, right?

 Aswathy said: (Sep 17, 2018) Hi, I am not getting this, Please explain this answer clearly.

 Yuvraj K said: (Jan 30, 2019) I think the Answer is 720.

 Raj Kumar said: (Aug 18, 2019) The answer is 360 because here in question 'e' is coming twice.

 Julius said: (Oct 9, 2019) In how many ways can the word obasanjo be arranged so that the vowels will never come together.

 Yaswanth said: (Nov 26, 2019) LEADER consist of 6 words, In that 2E's so, 6!/2! = 720/2 = 360.

 Rasi said: (Apr 2, 2021) Hey guys! LEADER = 6! =720 (E E) = 2! = 2 Now divide, 720/2 = 360. Now it is clear right?

 Sandeep Bodamwad said: (Jul 10, 2021) Yes, the answer 360 is very correct, because there letter E is repeated. So the whole letters count will be 5. 1+2+3+4+5+ = 360