# Aptitude - Permutation and Combination - Discussion

### Discussion :: Permutation and Combination - General Questions (Q.No.1)

1.

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

 [A]. 564 [B]. 645 [C]. 735 [D]. 756 [E]. None of these

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
 = 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2) 3 x 2 x 1 2 x 1
 = 525 + 7 x 6 x 5 x 6 + 7 x 6 3 x 2 x 1 2 x 1
= (525 + 210 + 21)
= 756.

 Naveen said: (Jul 12, 2010) First of all we need to select 3 man from 7 so = 7C3 = 35 Then we can choose 2 person from (7-3)+ 6 person ie. 10C2 = 45 So answer should be 53*45 = none of these.. Please tell me where I'm wrong.

 Nagu said: (Jul 15, 2010) Hi Naveen, you are wrong first step is correct that is First of all we need to select 3 man from 7 so = 7C3 = 35 and second step you are wrong in problem they are asking minimum of 3 men so already you took 3 men so now we need woman so we need 2 women from 6 women so 6c2 so the value of 6c2 is 15 now 35*15 = 525 like same way second committee we took 4 men from 7 men group i.e. 7c4 = 35 we need one more woman for 2nd committe because already we took 4 men so 6c1 = 6 so 35*6 = 210 now we took all men for 3 rd commitee so 7c5 = 21 now the total is 525+210+21 = 756 I hope you have understood the problem.

 Kumar said: (Dec 24, 2010) Nice answer nagu.

 Nikhil said: (Dec 27, 2010) Why would the logic 7C3 * 10C2 not work? First we've choosen 3 Men - 7C3. Now we have 10 people left (which included both men and women). Then we need to select any 2 people either men or women so 10C2. Please Clarify.

 Mohan said: (Jan 15, 2011) @Nikhil, The combinations in 7C3 and 10C2 are not mutually exclusive. Hence we cannot just multiply both and result in a solution. Its like saying, the combinations that 7C3 gives is same as what 7C1*6C1*5C1 would give.(selecting 1 out of 7 as first Man, 1 out of remaining 6 men for second...etc) That is 35 is not equal to 210.

 Madhusudan said: (Mar 14, 2011) In Above solution I am unable to understand why have we included 5/1(in both the committee and 6/2 in 2nd committee. Can any help me to understand how the combination considered in the example.

 Anu said: (May 14, 2011) 5 ball are to be placed in 12 boxes, they are placed in three rows such that each row contain at least one ball in how many ways it can be placed.

 Bhargav said: (May 29, 2011) Not a good answer to give please another method please easy method.

 Bhargav said: (May 29, 2011) How can you take a 3 men 2 women 4men and 1women and a 5 men please solve easy method.

 Rajeev said: (Jul 22, 2011) In the first step 7c5 converted 7c2 in the second step. But how ?

 Renuka said: (Aug 11, 2011) Because nCr = nCn-r.

 Kumamako said: (Aug 16, 2011) I think that it should be 3*2^3*3-3

 Lana said: (Aug 27, 2011) Hey can any one explain me how 210 came because I got 140 instead of 210. Lookin forward for your support.

 Kranthi said: (Oct 12, 2011) How 7C4 became 7C3 ?

 Siva said: (Oct 16, 2011) 7c4 = 7c(7-3) So came 7c3 @kranthi... Based on the formula nCr = nC(n-r).

 Leo said: (Oct 18, 2011) --> (7C3 x 6C2) + (7C4 x 6C1) + (7C5) Shouldn't it be like this? --> (7C3 x 6C2) x (7C4 x 6C1) x (7C5)

 Vivacity said: (Oct 21, 2011) @leo: It can be A or B or C. The basic rules of permutations so we have to add them up A+B+C...where A,B,C refers to the terms.

 Kanchan Srivastava said: (Nov 16, 2011) nCr = nCn-r and 7c4 = 35, 6c2= 15 I am unable to get it and how came 7c4 = 35 or 6c2= 15 Please explain it.

 Cyrus said: (Dec 20, 2011) nCr = nCn-r i.e n!/(r!*(n-r)!) So, 7C4 = 7!/(4!*(7-4)!) = 7*6*5*4*3*2*1/(4*3*2*1*(3*2*1)) = 7*6*5/3*2 = 35.

 Shiva said: (Dec 26, 2011) How do you find weather give problem can solved by combinations or permutation?

 Anshul Vyas said: (Jan 24, 2012) Mr. Mohan is absolutely right. We know that 7c3 is not equal to 7c1*6c1*5c1. In the same way we can't write 10c2 for rest two members.

 Ankz said: (Mar 14, 2012) I think nikhil's query still not answered Why would the logic 7C3 * 10C2 not work? First we've choosen 3 Men - 7C3. Now we have 10 people left (which included both men and women). Then we need to select any 2 people either men or women so 10C2. Please Clarify. Anyone please elaborate whats wrong in this approach ?

 Miguel said: (Apr 23, 2012) If nCr=nCn-r, then why isn't 7c3 (from 7c3 x 6c2) changed to 7c4?

 Abhishek said: (May 10, 2012) Is any shortcut solution of above question?

 Rishi said: (May 22, 2012) How 7c3= 35 ?

 Randika said: (Jul 15, 2012) What is meant by nCr=nCn-r ?

 Divya said: (Aug 25, 2012) May be 5 womans is also possible, why we don't take that option?

 Dipak said: (Sep 1, 2012) @divya because there is a condition that, 3 man's are compulsory has to be select..

 Vibhor said: (Sep 21, 2012) How to know when to use permutation and when to use combination ?

 Sujit said: (Nov 27, 2012) Permutation is used when asked to find number of arrangement and Combination is used to find number of ways of combination.

 Olga said: (Dec 16, 2012) I tried to calculate it differently. First, I calculated number of combinations with less than 3 men in them: 7C2*6C3=21*20=420. Then, total number of combinations possible: 13C5=1287. After that I deducted number of combinations with less than 3 men in them from total number of combinations: 1287-420= 862. However the answer is incorrect. Please, where is my mistake?

 Sri Ram said: (Dec 17, 2012) Is there any shortcut for doing the steps like 13c5 like that. These steps are taking too much time?

 Shiva said: (Dec 25, 2012) @Olga. You have removed only the condition where you have only two men, but you have to remove the condition where one men and no men are present. i.e. 7c2*6c3 + 7c1*6c4 +7c0*6c5 has to be removed from 13c5.

 Pankaj said: (Mar 3, 2013) Would you please explain in question they ask only for the three man combination then why you have taken 4m and 1m and all five men possibilities?

 Venki said: (Mar 10, 2013) Guys do you have any idea when to use permutations or combinations?

 Akshay said: (Apr 4, 2013) @Venki. If the order doesn't matter, it is a Combination. If the order does matter it is a Permutation.

 Anbu said: (Apr 18, 2013) @Pankaj. They asked at least 3 men not compulsory 3 men so 4men and 5 men is also possibility.

 Nishi Kant said: (Jul 11, 2013) Here it is clearly given that we have to select so here we have to apply the combination rule and at-least three is given so we have to start from three and proceed further.

 Riya said: (Aug 8, 2013) There are 4 types of permutation & its solved by diff formula. Which type is this?

 Shams said: (Sep 1, 2013) @Riya. This is a Problem of Combination, There is only one formula is applied that is, nCr = n!/(n-r)!r!. (3 men and 2 women). For man, n=7 & r=3. For women n=6 & r=2. (7C3 x 6C2) = 7! / (7-3)! 3! = 7*6*5*4*3! /4*3*2*1*3! = 840/24. = 35. So using this formula proceed further steps to get the result.

 Sweety said: (Sep 2, 2013) This method is ok but it will take much time to do in examination hall.

 Swetha said: (Sep 3, 2013) @Divya as in que asked as at least 3 men should be available in a group so we can't take 5 women. right?

 Sharon said: (Sep 13, 2013) GUYS. How do we analyse that whether permutation or combination has to be used here?

 H G said: (Sep 24, 2013) Permutations are used to find the number of arrangements and combinations are you used to find the number of different ways of groups.

 Rampal said: (Sep 29, 2013) Ya because they said to select at least 3 men which means we can select till 5.

 Shay Tee said: (Oct 21, 2013) Okay. How are 7c3 and 7c4 both equal to 35?

 Chithra said: (Nov 5, 2013) Combination is nothing but we are going to select from a particular group. In permutation also we will select from group, after selecting we need to arrange it according to the given condition. i.e.) Combination -> selection. Permutation -> selection followed by arrangement.

 Krishan said: (Nov 15, 2013) Anyone could explain why we used 7c2 in second step?

 Priyanka Das said: (Dec 8, 2013) 7C2 means 2 men are selected from 7 men.

 Deus Simon said: (Dec 17, 2013) The questions influencable but you did not talk about the of an event to occur.

 Akash said: (Jan 1, 2014) Can anyone please explain what does C imply here?

 Vatsal Khandelwal said: (Jan 2, 2014) Dear @Akash, You mustn't try these if you don't even know the meaning of "c". Else what I can say is that. "c" simply means "combination". But you can't understand if you haven't read the chapter.

 Ashok said: (Jan 5, 2014) Please give a correct explanation to combination and permutation?

 D.Yogapriya said: (Feb 24, 2014) Is there any other simple way to solve this 1st problem?

 Pandu said: (Feb 26, 2014) @Priyanka das. 7C2 has taken since in the last case we need to take 5 men so that you may be understood, its actually 7c5 when in nCr if n value is more than half of n value we need to use formula as nC(n-r)so implies n=7, r=5 since r is more than half of n we used as 7C(7-5) it becomes 7C2. OK I HOPE you got it.

 Ankit said: (Mar 16, 2014) Friends, please tell me how to identify whether question is of permutation or combination?

 Gokul said: (Mar 28, 2014) It is simple to identify the given problem is permutation or combination. It is just by common sense ya. In permutation, there is arrangement of things. But in combination it is selection. I hope you understand.

 Ales said: (Mar 29, 2014) Please can you tell me why we use combination instead of permutation?

 Pavz said: (Apr 8, 2014) You can't use 7C3*10C2. Because suppose the 5 men are ABCDE and 2 women are XY. If you say 7C3*10C2 let the men be ABC and the other two be DX then the committee will be ABCDX. If you consider another way for 7C3*10C2 let da men be ABD and the other two be CX. Now the committee is ABDCX. But ABCDX = ABDCX. But by using 7C3*10C5 it is taken as two separate ways.. but it is one. So that method is wrong :).

 Vamsi Reddy 1100 said: (Apr 25, 2014) Hey guys! I am Vamsi I got a small doubt. Please clear my doubt. For example if we want to choose 3 items from a list of 10 items, can the answer be c(10,1)*c(9,1)*c(8,1) instead of c(10,3). My interpretation was I broke down the process of choosing into choosing 1 thing for 3 times...and then apply product rule. Please help me.

 Santosh said: (Jul 12, 2014) It is so long procedure, is any shortcut method for this?

 Kanaga said: (Jul 22, 2014) Which formula using this method clearly?

 Ankit said: (Aug 3, 2014) @Vamsi. As @Mohan said to do multiplication A * C for instance event A shld not in any way affect the occurrence of C. Now from a list of 10 if 1 item is selected then that item can never be selected from the list of next 9 items AND HERE IS THE CATCH so removing any specific item in c(10,1) will make sure that specific item will not be selected in c(9,1). So c(10,1) and c(9,1) is not mutually exclusive since they both remove elements from same list and WE CAN NOT MULTIPLY NON MUTUALLY EXCLUSIVE events. Let assume, there are three nodes STATION 1 , STATION 2, STATION 3. If we have 3 path from STATION 1 to STATION 2 and 3 path from STATION 2 to STATION3 then we have 9 path from station 1 to station3, but think if taking any from station 1 to station 2 causes blockage of any one path from station 2 to station 3. Then total number of path from station 1 to station 3 is 5. Hope you got it :).

 Vijay said: (Sep 11, 2014) Hi, Could you please explain in what condition or scenario the formula nCr=nCn-r will implicated. Its used in (7C3 x 6C2) + (7C4 x 6C1) + (7C5) = ......... + (7C3 x 6C1) + (7C2)ncr= ncn-r With this formula I came to know that 7c4 and 7c5 will become 7C3 and 7c2 respectively. But I could not understand why this formula is used here, Why cant we use the formula ncr=n!/r!(n-r)!. Pleas anybody kindly help me in clearing the above doubt. Thanks in advance.

 Sagar said: (Oct 21, 2014) I am repeating @Nikhil and @Ankzs and my query here: I think @Nikhil's query still not answered Why would the logic 7C3 * 10C2 not work? First we've choosen 3 Men - 7C3. Now we have 10 people left (which included both men and women). Then we need to select any 2 people either men or women so (6+4=10)C2. Please Clarify. Anyone please elaborate what's wrong in this approach ? Why do we need to separately consider cases in which there are 3 men, 4 men and 5 men? 10C2, [ having 4 men and 6 women ] includes the possibilities of selection of 4 guys and 5 guys.

 Rakesh said: (Nov 4, 2014) = (7C3 x 6C2) + (7C4 x 6C1) + (7C5). = 7 x 6 x 5x6 x 5+ (7C3 x 6C1) + (7C2). 3 x 2 x 1 2 x 1. From 1st to 2nd line I have not understood please ex-plane. How (7C4 x 6C1) + (7C5) become (7C3 x 6C1) + (7C2)?

 Prashant Olekar said: (Nov 7, 2014) According to combination rule we can write 7c3 as (7*6*5/3*2*1). Similarly. = (7c3*6c2) + (7c4*6c1) + (7c5). = ((7*6*5/3*3*1)*(6*5/2*1)) + ((7*6*5*4/4*3*2*1)*(6)) + ((7*6*5*4*3/5*4*3*2*1)). = 525+210+21. = 756.

 Bhawani said: (Jan 14, 2015) I have a small doubt. Why we didn't divide it like this 7c3*6c2+7c4*6c1+7c5*6c0/13c5?

 Mayur said: (Mar 11, 2015) Why only that formula used in 7c4 n 7c5. ncr=nc (n-r)? Why didn't we use in 6c1? Can anybody clear me please?

 Ananya said: (Mar 16, 2015) Is there any shortcut for this?

 Naji said: (Mar 22, 2015) How can we notice that we just have to select without arrangement ?

 Sam said: (Jun 4, 2015) Can you please explain 7c5 is converted into nc(n-r)? But why didn't convert 6c2 into 6c(6-2)?

 Naveen Kumar said: (Jun 17, 2015) Why we are calculating 7C5 as 7C2?

 Maryann said: (Aug 9, 2015) Can someone explain better for me because I'm getting confused?

 Ajay said: (Aug 17, 2015) 7C5 written as 7C2 because 7C5 = 7C(7-5) both having same values from formula nCr = nC(nr).

 Ajaykumar said: (Aug 17, 2015) 7C3 = 35, 7!/(7-3)!3! = 7!/4!3! = 7*6*5/3*2*1 = 35.

 Y.Deepikamanoj said: (Oct 28, 2015) How many 3 digit numbers can be formed from 2, 3, 4, 5, 6, 7, 9, which are divisible by 5 and none of the digits is repeated? Please tell answer and method for this problem.

 Zara said: (Nov 10, 2015) I used combination formula, nCr = n!/r! (n-r)! Required number of ways = (7C3*6C2)+(7C4*6C1)+7C5. = (7*6*5/3*2*1+6*5*1/2*1)+(7*6*5*4 /4*3*2*1+6/1)+(7*6*5*4*3*2*1/5*4*3*2*1). = 35*15+35*6+24 = 756 answer.

 Trinity said: (Nov 23, 2015) Why 7C3*10C2 is wrong? Lets take a case: We have 7 men namely A B C D E F G and 6 women H I J K L M. Case 1) 3 men and 2 from the set of 10 = A B C + D H. Case 2) 3 men and 2 from the set of 10 = A B D + C H. Both the sets are identical. This shows that our sets are not mutually exclusive. Hence we must choose men and women separately.

 Akshay said: (Jan 16, 2016) How 6C2 equal to 15 please explain?

 Eswaru said: (Feb 2, 2016) In that question that is cleared that 5 members need to be select out of 7 men and 6 women. And in question there is a condition that at least 3 men that mean we can select in 3 ways that we can select 3 men or 4 men or 5 men. 1) 3 men and we want another 5-3 = 2 women. We have to select 3 men from 7 men that is 7C3 and 2 womens from 6 women that is 6C2 results is 7C3*6C2. 2) 4men and we want another 5-4 = 1 women. We have to select 4 men from 7 men that is 7C4 and 1 women from 6 women that is 6C1 results is 7C4*6C1. 3) 5 men and we want another 5-5 = 0 women. We have to select 5 men from 7 men that is 7C5 and 0 women from 6 women that is 6C2 results is 7C5*6C0. Answer is we need to sum up all these combinations that is (7C3*6C2+7C4*6C1+7C5*6C0) that is 756.

 Daniel said: (Feb 11, 2016) I am still have little problem with the solving.

 Daniel said: (Feb 11, 2016) I am still have little problem with the solving.

 Malu said: (May 15, 2016) Why can't we select only five women out of 6 women. Why that combination is not taken into account.

 Ishrat Jaleel Khan said: (Jul 10, 2016) The question is not specific regarding the selection of men and women after the selection of at least three men. If we take the method [7C3 * 6C2] + [7C4 * 6C1] + [7C5 * 6C0] as the solution, then we must understand that all men are taken as one entity first then women are considered as another entity. Hence, 3 men are one entity and 2 women are another entity. Similarly, 4 men are one and likewise, 5 men are one entity. That means if you are done with the selection of 3 men at least, then DONT SELECT MEN AGAIN from the remaining. And if you choose to select men again, THEN INCLUDE THEM WHILE SELECTING ALTEAST THREE MEN.

 Simran said: (Jul 28, 2016) I understood all the combination how we choose, Thanks for this @ Mohan & @Pavz. :) But I have little doubt in solving cases like; 1) why we "multiple" in case? 2) why we "add" all cases?

 Anil said: (Aug 31, 2016) @Malu. The problem says at least 3 men so if you select 5 women it would contradiction to the problem.

 Fatmata Sorie said: (Sep 30, 2016) Where do you get from 35?

 Kiprono Langat Esau said: (Oct 25, 2016) Kindly, solve this. The committee of six is to be formed from a group of seven engineers and four mathematicians, how many different committees can be formed if at least two mathematicians are always to be included?

 Siva Kumar said: (Nov 19, 2016) Nice solution @Nagu.

 Ishak said: (Jan 8, 2017) Remarked first line 7c5 second line 7c2. How?

 Abhimanyu said: (Feb 4, 2017) 756 WILL BE CORRECT. 35 * 15 + 35 * 6 + 21= 756.

 Rutuja said: (Feb 7, 2017) Someone, please give the solution. In a group of 6 people, there are 3 Indians and 3 Chinese. How many subsets can be created such that there are at least 1 Indian in each subset?

 Harsh said: (Mar 8, 2017) It's correct that we select 3 men from 7 and 2 women from 6. But why do we need to multiply it to get total no of ways?

 Shoeib said: (Mar 10, 2017) Yeah, I too agree @Harsh.

 Ali said: (Apr 26, 2017) @Harsh. I think we need to multiply so that we get the total number of ways we can choose 3 men from a total of 13 men and women.

 Mamta Patel said: (Jun 14, 2017) The total number of combinations possible: 13C5=1287. Then Calculate the number of combinations with less than 3 men in them: 7C2*6C3=21*20=420 7C1*6C4=7*15=105 7C0*6C5=1*6=6. After that deduct the number of combinations with less than 3 men from total number of combinations: 1287-(420+105+6)= 756.

 Felix said: (Jun 16, 2017) How is 7c3 = 35?

 Aryabhatt said: (Jun 17, 2017) Why don't we apply permutation after getting the answer 756 to get the number of ways those 5 persons can be arranged?

 Naina said: (Jul 25, 2017) What are the number of ways to select 3 men and 2 women such that one man and one woman are always selected? The Answer is 30 and solution is 4c2 * 5c1. But, I don't think it is right according to the condition given in the question. Someone help me.

 Riya said: (Aug 2, 2017) I did not understand how we got 21 for 7C5?

 Abhijit said: (Sep 14, 2017) I tried to solve like this. Out of 7 places first 3 places to be filled by men so.. 7*6*5 = 42*5 ways but order does not matter so removing redundancies (42*5)/3!. Similarly Remaining two places can be filled in (10*9)/2!. Final answer is same as 7C3 * 10C2.

 Tshewang Chophel said: (Oct 13, 2017) How to get 7c3=35?

 Sneha said: (Oct 30, 2017) Any one know the another method of solving?

 Udoy said: (Nov 9, 2017) When do we add or multiply ?

 Manuel said: (Nov 9, 2017) Help me knowing how to solve this kind of equation I'm doing my Student Individual Learning Plan. I need to proof that I was correct so help me please.

 Calesto said: (Nov 12, 2017) I don't understand why 7C5, in which 5is total number of persons to be chosen where 3 are men and remaining two are obviously women. Thus, the answer should be 735.

 Teja said: (Mar 9, 2018) Here, ncr=nc(n-r).

 Pratheeksha said: (Mar 24, 2018) Yes, 756 is a right ans. Bcause out of 13(7men+6women) we have to select 5.Therefore 7men(5)6women, 7C3 X 6C2=35(15)=525, 7C4 X 6C1=35(6)=210, 7C5 X 6C0=21(1)=21, =756. Therefore, 756 is the answer.

 Vamsi Mvk said: (Jun 18, 2018) @Pavz. Thanks for your explanation.

 Dimi said: (Jun 22, 2018) I have a question, in how many ways can a committee of nine people can be formed from ten men and their wives. If no man is to serve on the committee with his wife? Please solve this.

 Shirin Sultana said: (Aug 13, 2018) But many ones falling problems not finding the details solution of the combination. It might be done more details as like: 7C4 = 7!/4!*(7-4)! = 7*6*5*4!/4!*3*2*1. = 7*6*5/3*2*1, = 35. And that would be understandable to all. By the by, May I know the basic differences between combination and permutation.

 Dadasaheb Maske said: (Aug 18, 2018) Among 5 children there are 2siblings.in how many ways children be seated in a row so that the siblings do not sit together. A) 86 B) 72 C) 46 D) 38 How to solve it? Please anyone explain me.

 Aishwarya said: (Aug 21, 2018) @Dadasaheb. The total ways-5!=120. if siblings sit together-2!4!=48. Hence; 120-48=72.

 Ishika Singh said: (Aug 29, 2018) I am not getting the solution. Please, anyone explain to me.

 Tanvir said: (Nov 21, 2018) Basic Formula is nPr=n!/(n-r)! n=number r=spaces And in case of repetition(if number/character is repeating) we have to apply this formula :nPr/(No .of repeated number/character)! ==>n!/(n-r)!/(No .of repeated number/character)! Based On Question : We got three possibilities: First possibility:- (7 men 6 women) out of which we can select (3 men 2 women) =>>(7P3/3! * 6P2/2!) ==>(35*15) {men repeated 3 times and women 2 times that's why i have take 3! and 2!} ==>525----> (1) Second possibility:-(7 men 6 women) out of which we can select (4 men 1 women) ==>(7P4/4! * 6P1/1!)==>35*6 = 210-----> (2) Third possibility;-(7 men:6) out of which we can select (5 men only ) ==>7P5/5! = 21----> (3) Add all equatations (1),(2) and (3) ==>525 + 210 + 21 = 756.

 Dagwidhi said: (Feb 23, 2019) Why can't we do this? Select 3 out of 7 men. Because that's the main condition. Then select 2 women out of 6/1 man out of 4 and 1 woman out of 6/ 2 men out of 4. i.e 7c3 (4c2 + 6c2 + 4c1. 6c1)? Please tell me.

 Anand Kumar Gupta said: (Mar 26, 2019) Here 5 persons are select, and at least 3 men including this committee so we have Men. Women 7c3 * 6c2 = 525 7c4 * 6c1 = 210 7c5 * 6c0 = 021 = 756 is the answer.

 Vinay Teja said: (Jul 30, 2019) There are 5members required so why don't we divide with 11c5?

 Nivetha said: (Aug 24, 2019) Shouldn't we have to divide the whole answer by 13c5? Please tell me.

 Akshay said: (Nov 13, 2019) Yes it's write answer. Using combination formula. nCr = n! / r! ( n - r ) ! Question says at least 3 men's , means we have to select minimum 3 men's , it means we can select either 3 men's or 4 men's or 5 men's. 1)in first case select 3 men's and 2 women's. To form 5 people commette .. 7! / 3! (7-3)! * 6! / 2!(6-2)! - Eq-1 2 ) in second case select 4 men's and 1 women can be selected in. 7! / 4!(7-4)! * 6! / 1!(6-1)! - Eq -2 3) in third cas we can select 5 men's only Can be selected in 7! / 5!(7-5)! - Eq -3 The final answer is: Eq-1 + Eq-2 + Eq-3= 7! / 3! (7-3)! * 6! / 2!(6-2)! + 7! / 4!(7-4)! * 6! / 1!(6-1)! + 7! / 5!(7-5)! = 525 + 210 + 21 =756 ways.

 Sake said: (Jun 16, 2021) Why can't we do it like this, Total ways = at least 3 men + not men. i.e not men = 6c5. Total ways= 13c5. therefore, 13c5 - 6c5.

 Shaifali said: (Jul 9, 2021) Well explained @Nagu. But I don't understand why do we need to multiply? why can't we add?

 Jay said: (Sep 10, 2021) @Nivetha. We only have to find total combinations, not the probability. If we want to find probability then we have to divide it by 13C5.