# Aptitude - Permutation and Combination - Discussion

### Discussion :: Permutation and Combination - General Questions (Q.No.4)

4.

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

 [A]. 210 [B]. 1050 [C]. 25200 [D]. 21400 [E]. None of these

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= (7C3 x 4C2)
 = 7 x 6 x 5 x 4 x 3 3 x 2 x 1 2 x 1
= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

 Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120.

Required number of ways = (210 x 120) = 25200.

Video Explanation: https://youtu.be/dm-8T8Si5lg

 Anil said: (Oct 4, 2010) Why multiply with 5?

 Rahul said: (Oct 28, 2010) @ Anil: The basic idea here is that first you pick out 3 consonants out of 7 and 2 vowels out of 4. Then you take the 5 letters that you have and make a word out of it. To pick out 3 consonants out of 7 you use the combination without repetition formula which states n!/(n-r)!r! which gives you 7C3 and the same for vowels gives you 4C2. Once you have done that, you still need to make a word of it. You have 5 letters now that can be arranged into 5! ways to make a word so you multiply the three values since each step is dependent on the step before to make the final word. The answer comes out to be 35 x 6 x 120 respectively for each step mentioned chronologically.

 Student said: (Mar 18, 2011) Can't we apply permutation directly here since in this question arrangement of words is to be done. i.e. 7P3*4P2 = 2520 One zero less to match the answer of 25200 :(

 Pranay said: (Jul 18, 2011) The question doesn't mention "distinct" consonants or vowels. So repetition should be allowed.

 Suchi said: (Aug 9, 2011) c = consonants v = vowels Why has the num of letters has to be 5? Can't it be 4c+3v(7 letter) or 5c+4v(9 letters) Then how can v take 5! ?

 Machender said: (Aug 22, 2011) Miss suchi, from given question it was clear that out of 7 consonants 3 consonants and 2 vowels of 4 vowels could considered.

 M.V.Krishna/Palvoncha said: (Sep 10, 2011) Hi suchi. Given 3c out of 7c 2v out of 4v. So, required number of letters are 5. they can be arranged among themselves in 5!. Hope you understood.

 Student said: (Nov 1, 2011) Why we need multiply 120 with 210 is it necessary?

 Navin said: (Dec 18, 2011) Hey they anly asked us how many ways to form the letters they didint ask how many ways to arrange it the answer is wrong it should be 210.

 Sachin Jain said: (Jan 7, 2012) Dear friends Keep in mind first that it is a permutation problem becoz the order of letters matter in order to make different words. Secondly out of 7 consonants we have to select 3 so when we select a consonant out of 7 it will not again be considered to be selected so the repetition is not allowed. Now the formula for Permutation without repetition is (n!)/(n-r)!. So it will be 7P3*4P2. =(7!/4!)*(4!/2!)=2520 (not 25200).

 Litu said: (Mar 9, 2012) Hi, I do agree with Sachin Jain as it is a permutation query. We can arrange it in any way using using 3 consonants out of 7 & 2 vowels out of 4. Ex: If there is bcdfghj then we can do it bcd n also cbd so it is permutation type not combination so the answer should be 7P3 * 4P2 = 2520 But again we have to arrange these 2 groups together. So I'm not getting form this point. Can anyone clear the same?

 Rizwana said: (Jun 21, 2012) Hi, I always find problem in using C and P in formula I could not distinguish when to use nCr and when to use nPr please help in finding difference.

 Benni said: (Jun 25, 2012) How can we find permutation and combination problem ?

 Veyron999 said: (Aug 31, 2012) This question can be solved by two methods 1) using combinations: we have to select 3 consonants from 7 and 2 vowels from 4, we will use 7C3 x 4C2, this implies that for eg. consonants are B C D F G H J, then if BCD is there then CBD and DBC won't be there and so on.. but they will make different words, so we arrange them later on by multiplying by 5! .. eg. one combination will be BCD AE this can be arranged in 5! ways and so on... so ans. is 7C3 x 4C2 x 5! = 25200 2) using permutations: unlike previous method if we use 7P3 we also consider BCD, CBD and all possible arrangements... then we get combinations of the type CCCVV (C- consonant, v- vowel) but they can be arranged in 5! ways eg- CVCVC, VVCCC, etc.. but using permutation, we have already accounted for those combinations in which all 3 consonants occur together and 2 vowels occur together.. so to eliminate those cases, we need to divide it by the repetitions i.e. divide by 3!2! (for consonants and 2 for vowels) and then multiply this with the initial result.. so ans is 7P3 x 4P2 x 5!/(2!3!) = 25200 tried to explain as clearly as possible..hope i was of some help..

 Prasant said: (Mar 5, 2013) Finally multiplied with 5!, because, each word is consisted of 5 distinct letters. Had it been persons in place on letters, then 5! multiplication would not have been applied.

 Math Wiz said: (Apr 21, 2013) NO. It's not because of the letters (and not person) we should multiply with 5!, but it's about what the question said; how many words can be FORMED? that's we call PERMUTATION. At first we might have thought it's just COMBINATION because we didn't notice the last phrase of the question said how many words can be FORMED. So it's considered as both PERMUTATION and COMBINATION.

 Shub said: (May 9, 2013) I don't know why all are confusing as permutation and combination. I think this question"how many words can be FORMED " is clearly combination.

 Harsha said: (Sep 1, 2013) When I saw this question I was like it will be permutation since arrangement of words is important and I did 7P3*4P2(just like other sums). Why does the method differ here !

 Shams said: (Sep 3, 2013) @Harsha. Here the asking question is group of words (use combination) not to asking about the order of words so don't use the permutation. OK.

 Harsha Vardhan said: (Sep 4, 2013) Here we should use permutations we need to use combinations and just selecting the consonants and vowels from the given 7 consonants and 4 vowels, its not necessary to multiply with 120 there so according to me, its 210 just selecting those consonants and vowels from the given one not more than that.

 Akhila said: (Nov 10, 2013) How can we find the given problem is from permutation or combination?

 Kaustav said: (Apr 9, 2014) Combination 3 consonant and 2 vowels is still left. By adding forms 5! and then rest to find the total.

 Jyoti Bala said: (Jun 23, 2014) When we apply permutation and combination? please tell me one.

 Uday Sai Ranjan said: (Jun 24, 2014) We can do this problem with permutation. Let me explain. We got 2520 by permutation but its not complete. It is only the permutation of two sets of alphabets. So what we have missed out is combining both the sets and the number of ways of combining the two sets (3c and 2v) is 10. So finally we get 25200 as the answer. They are: vvccc vcvcc vccvc vcccv cvvcc cvcvc cvccv ccvvc ccvcv cccvv

 Jatin Bavishi said: (Aug 18, 2014) This cannot be a permutation problem because the first thing we ought to do is select 4 consonants out of 7 and 2 vowels out of 4. Let us approach the question by assigning actual alphabets, and reducing the quantity. Suppose there are 3 consonants B, C and D and 2 vowels A and E, and we have to form a word using 2 consonants and 1 vowel. The required words are: ABC, BAC, BCA, ACB, CAB, CBA, ACD, CAD, CDA, DCA, DAC, ADC.....the sum would be 36. Let us now do the sum using formula: 3P2*2P1=12, which is clearly not the answer.

 Sagar said: (Sep 23, 2014) If it was mentioned distinct in the question can we use 7p3*4p2.

 Munjal said: (Oct 11, 2014) Difficult question : First of all it is not the question of combination, it is of permutation (assumed : letters without repetition). Reason : suppose there are two letters I want to arrange le t say they are t and p, so tp and pt conveys the different words. So order matters here, so we use 7P3*4P2 instead of 7C3*4C2. So the answer is 5!*7P3*4P2.

 Achu said: (Dec 26, 2014) What's the need to take 5 here? They didn't mention any groupings here. Then what's the purpose to group?

 Mansi said: (Jan 4, 2015) What is the difference between permutation and combination? I always get confused.

 Ridsie said: (May 5, 2015) A bag contains a total at 105 coins of Re.1, 50 paise and 25 paise denominations. Find the total number of coins of Re.1, if there are total number of 50.50 rupees in the bag and it is known that the number of 25 paise coins is 133.33% more than Re.1 coins.

 Samiksha said: (May 8, 2015) @Ridsie. Let the no. of Re. 1 coin be x. ==>no. of 25 paise coins = x+1.33 x = 2.33 x. ==>no. of 50 paise coins = 105-(x+2.33 x) = 105-3.33 x. (1)x+(25*2.33 x)/60+(50*(105-3.33 x))/60 = 50.50. Solve for x.

 Spurthy said: (Jun 24, 2015) I cannot get you, after what you said after 120 ways?

 Neha said: (Jun 30, 2015) Why we multiply with 5! I cant understand. If we multiply with 5! in words then why we not multiply in persons it will also formed in different manner. Please someone explain this.

 Ajay said: (Jul 1, 2015) Can anyone help me to understand when have to take NCR or when have to take NPR in easily manner. I am more confused about this concept.

 David said: (Sep 4, 2015) Simplify problem. 3 consonants BCD. 3 vowels AEI. Choose 2 consonants, no repetition. Choose 2 vowels, no repetition. Form different words from the chosen consonants and vowels. Choose 2 consonants, 3 ways - BC, BD, CD. Choose 2 vowels, 3 ways - AE, AI, EI. Different groups with 2 consonants and 2 vowels, 3x3 = 9. BCAE, BCAI, BCEI, BDAI, BDAI, BDEI, CDAE, CDAI, CDEI. Different words from 4 character set such as BCAE, 4! = 4x3x2x1 = 24. BCAE, BCEA, BACE, BAEC, BECA, BEAC. CAEB, CABE, CEAB, CEBA, CBAE, CBEA. ABCE, ABEC, AEBC, AECB, ACBE, ACEB. EBCA, EBAC, ECBA, ECAB, EABC, EACB. Total 4 character words from 2 different consonants and 2 different vowels would be 9x24 = 216. Apply to above. 7C3 x 4C2 x 5! = 25200. More difficult would be when repetitions are allowed in consonants and vowels.

 Rob_Ph said: (Oct 5, 2015) 210 is correct. What is asked in the problem is how many words that have 3 consonants and 2 vowels can be formed from the given 7 and 4. Basically all words that contains 3 consonants no matter how you arranged the 3 consonants that you picked is considered as one, same with the vowels. Meaning QEWAR is same with WEQAR, hence counted as 1 sample. Unless you choose another consonant say QEWAD or another vowel QIWAR then it will be a different sample.

 Athar said: (Oct 7, 2015) You have to arrange the letter. 3 consonants out of 7. 2 vowels out of 4. 7c3x4c2 = 210. You can do, How many letter out of these letters? 3+2 = 5. Now the required no of ways = 210 x 5!

 Leonardo said: (Jan 30, 2016) I don't understand it.

 Rishi said: (Mar 25, 2016) In question instead of how many words there should be how many arrangements. Otherwise, a solution is wrong according to question.

 Prof. Vedavyas said: (May 12, 2016) Explanation is actually simple. Suppose if they asked to find the number of words possible using five different letters and you had no constraints of consonants and vowels. Then the answer would obviously 5!. So, we know you can form 5! words from five letters. Now, in the problem, there are not just 5 letters, but much more - 7 consonants and 4 vowels. The 3 consonants can be selected in 7C3 ways. For each of these, we can select 2 vowels in 4C2 ways. So, the 5 letters themselves can be selected in 7C3X4C2 ways. Note that each of these 7C3X4C2 are a different set of five letters. A given set of 5 letters can be arranged in 5! ways. So, 7C3X4C2 sets of 5 letters can be arranged in 7C3X4C2X5! ways. Hence the answer is 25,200

 Atul Moundekar said: (Jun 3, 2016) This question is only based on a combination, they never told to find the arrangements. So the answer is 210.

 Mahima said: (Sep 10, 2016) Why can't we use permutation here instead of combination and multiplication by arrangement? Since permutation = combination * arrangement. But the answer is coming 2520.

 Nitish said: (Nov 4, 2016) As the consonants and vowels are not specified and we only have to find how many ways we can arrange them i.e. 210 the Order does not matter. If the letters were specified then only we can apply permutation here. For example, if the question was 7 men and 4 women, we wouldn't multiply with 5!. Hence, the correct answer would be 210.

 Sundeep said: (Nov 7, 2016) @Veyron999. Best explanation.

 Sandhya said: (Dec 17, 2016) 3 out of 7 consonants, 7 * 6 * 5/3 * 2 * 1 = 210/6 2 out of 4 vowels 4 * 3/2 * 1 = 12/2 = 6, 210/6 * 6 = 210. Formation of 3 Consonants and 2 vowels 3 + 2 = 5. 5 * 4 * 3 * 2 * 1 = 120, 210 * 120 = 25200.

 Mohan said: (Dec 31, 2016) It's. VCVCC VCVCV VCCVC VCCCV CVVCC CCVVC CCVVC CCVCV CCcvv

 Aditya said: (Mar 31, 2017) It's simple, Select 3 consonants out of B,C,D,F,G,H,J (assume these 7 consonants) in 7P3 ways. So, [ B C D ] is one selection out of 7P3 ways right? but, [ C B D] , [ C D B ] , [ B D C ] , [ D C B ] , [ D B C ] , all these contain same consonants. This is why you use the combination, to form groups of 3 consonants out of 7 consonants. Same goes for vowels : 4P2 ways to get groups of 2 vowels out of 4 vowels. total groups of consonants and vowels : 7P3 * 4P2. Now, we need to form words of 5 letters consisting of vowels and consonants. How many ways? In 5! ways. Thus, 7P3 * 4P2 * 5! = 25200 ways.

 Hehe said: (May 2, 2017) I don't think that it's supposed to be multiplied by 5! since it doesn't ask for the arrangement of the letters.

 Neeraj Tailor said: (Jun 26, 2017) How many numbers greater than 5000 can be formed with the digits, 3, 5, 7, 8, 9 no digit being repeated? In this question we won't multiply instead of adding?

 Himatheja said: (Aug 18, 2017) 4C2 = (3*4)/(2! * 2!). 2 VOWELS OUT OF 4. then, why 210?

 Ankita said: (Aug 27, 2017) @Himatheja. We have to make word from both consonants and vowel that's why we have to multiply combination of consonants with vowel also, 7c3*4c2. = 35*6. = 210.

 Aadesh said: (Sep 2, 2017) I agree @Ankita. That as we are picking up the words by applying combination means that we are forming the words therefore answer. It Should be 210.

 Probodh said: (Sep 5, 2017) The correct answer should be option A.

 Adithya Un said: (Oct 2, 2017) Here, the answer is 302400. Because you should 3 constants from 7 and 2 vowels from 4 and multiply then the answer is (7*6*5) * (4*3). Later you should arrange them by multiplying with 5!.

 Burt Yellin said: (Nov 5, 2017) I'm not sure this is the correct forum for this question, but here goes. If I have 4 consonants and there are 5 vowel sounds that can be pronounced hard or soft, how many different pronunciations are possible? If this is the wrong forum can you direct me to a place that could answer that question?

 Gezoi said: (Dec 21, 2017) 7P3 * 5P2 =2520. Then considering combining them accounting for repetitions we get = 5!/(3!*2!)=10. Therefore =2520*10=25200.

 Mark Burnham said: (Dec 23, 2017) This is a question of permutation, as the order in which letters are arranged changes the final result. Therefore = 7P3*4P2 = 2520. However, there are 10 different ways that the consonants and vowels can be arranged. So, the answer is 2520*10, =2520.

 May said: (Mar 20, 2018) Why do 5!? What if the consonants or vowels have repetition? Does it necessary to multiply with 5!?

 K Anirudh said: (Mar 24, 2018) Hi, @All. I am just confused with committee problems and this question in committee problems we are not doing this 5! thing if a committee of 5 people is to be made, we are just selecting the men and women out of given men and women and multiplying why it is not similar in this case. Can anyone help to understand the problem? Please help me.

 Rahul said: (Dec 1, 2018) Why we want to multiply with 5! Please tell me.

 Mihul said: (Apr 6, 2019) Why at 7p3 4 is excluded can anyone help me with this?

 Liam Martin said: (May 6, 2019) Here we can directly permute 7p3*4p2=2520 so this gives us 3 vowels say "aei" and 2 consonants say bc so aeibc we can then calculate the number of groups made by the 2 consonants or 3 values from this group of 5 letters chosen in total so the solution is 7p3*4p2*5c2 = 7*6*5*4*3*(5*4)/2=2520 *10 = 25200.

 Bo Zhang said: (Aug 31, 2021) @All. Here, it should be done with permutation at the beginning instead of using combinatation, which just make no sense. The calculation starts from the permutation of 7!/4! * 4!/2!. Once this is finished, we still have to figure out a way to arrange these 5 letters. We can start from the simplest pattern, which is 3 consonants next to each other on the right side and 2 vowels next to each other on the left side. Since we have already worked out the all the possible changes for every single letter in the previous step, we can know start to work on how many different patterns these five letters can be possbily arranged. There is 4 arrangements if the the first letter is vowel and 3 arrangements if the second letter is vowel. (the one that is already observed does not count). Eventually, you will get 1+2+3+4=10 different sets of letters. You should then mutiply the result from the perious permutation caculation (2520) with 10, because each set of letters still has 2520 different arrangements. The result will finally be: 7!/4! * 4!/2! * (1+2+3+4）=25200，. I hope, this will calrify the confusion.