### Discussion :: Permutation and Combination - General Questions (Q.No.6)

Pri said: (Jan 4, 2011) | |

I'm a bit confused. In 1st step : ... + (6c1 x 4c3) + ... In 2nd step : ... + (6c1 x 4c1) + ... How ? |

Aditya said: (Jan 7, 2011) | |

Hi Pri, Its because nCr = nC(n-r). So 4C3 = 4C(4-3) = 4C1. Its one of the formulaes. |

Phanichander said: (Feb 24, 2011) | |

Hi it is simple, read question ones again We have 6-boys and 4-girls we have to select only 4 members from the groups so in that once condition least 1 boy means one boy or more 1 than boy or all the boys only (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)=209 |

Udaya said: (Mar 28, 2011) | |

This question can be solved in a more simpler way . The requirement is that there should be atleast one boy out of the 4 children selected. So we get the output by subtracting the number of ways in which no boy is selected from the number of ways in which 4 children can be selected from total(i.e,10) Here goes the solution : Total no. of children(including boys n girls) = 10 The number of ways in which 4 children can be selected from 10 children is = 10C4 The no. of ways in which no boy is selected = 4C4 So the number of ways the selection can be made such that atleast one boy should be there is = 10C4-4C4=209. As simple as that ...Rite :) |

Jessie said: (Apr 10, 2011) | |

Udaya, your answer was really simpler. |

Abhishek said: (Jun 10, 2011) | |

Udaya its easy. But its more calculative. |

Amit said: (Aug 13, 2011) | |

It is very simple. Only read carefully after that ans we get. |

Richa said: (Aug 19, 2011) | |

It can also be done in another way... we can calculate total no of ways to select the children..ie 6+4=10 10C4 and subtract ways in which no boy is there ie all girls 4C4... 10C4-4C4= 209 Calculation would be easy in this way |

Thiyanesh said: (Sep 19, 2011) | |

In second step Why we use the nCr formula only for 4C3 and 6C4? |

Anukul said: (Oct 16, 2011) | |

Hey Thiyanesh you are little bit confused, read the question again. |

Joe said: (Oct 28, 2011) | |

1 boy can be selected in 6 ways. The rest 3 can be selected from the remaining 9, in 9C3 ways=84 ways. Hence the total no: of ways = 6*84 = 504. Is it right? |

Pallavi said: (Nov 10, 2011) | |

Another easy method is 1-(4C4/10C4) =1-(1/210) =209/210 Therefore 209 times out of possible 210. |

Kanchan said: (Nov 16, 2011) | |

4C4 How we get value by it? Means what it is? What is the meaning of it? |

Geet said: (Dec 6, 2011) | |

How to use ncr=ncn-r formula?why it is not used in second number 6c2 x 4c2? |

Msrinivasarao said: (Dec 14, 2011) | |

We have 6-boys and 4-girls we have to select only 4 members from the group In that at least one boy should be there.so we cab write total number of ways-no boy can be selected 10c4-4c4=210-1 =209 |

Djh said: (Feb 3, 2012) | |

@joe Your method is wrong because you cannot multiply 6C1*9C3 as it is a basic contradiction of m*n rule. The multiplication of m*n is possible only if they are the respective no of possibilities of two separate stages for the outcome of an event. Here the first stage is boys(6c1) and the second stage will be both boys+girls(9C3). Hence m*n cannot be done here as the second event is not a separate(girls only)stage. Hope I put it across clearly. |

Kaushik said: (Mar 7, 2012) | |

Thank you Aditya..! |

Jyoti Nagpal said: (Apr 18, 2012) | |

Another Shortcut Method: Total student=6+4=10 The number of ways to selecting 4 girls=10c4, The number of ways to selecting no boys= 4c4 because total boys are=6 so 10-6=4 to select 4 childern so, 4c4 so selecting atleast one boy 10c4-4c4=210-1=209. |

Jigs said: (Jun 15, 2012) | |

Another Shortcut Method:. Total student=6+4=10 The number of ways to selecting 4 girls=10c4, The number of ways to selecting no boys= 4c4 because total boys are=6 so 10-6=4 to select 4 children so, 4c4 so selecting at-least one boy 10c4-4c4=210-1=209. |

Ash said: (Jun 26, 2012) | |

Why nCr formula is not used here (6C2 x 4C2) , (6C3 x 4C1).? Can any one explain me this |

Shanmuga Kumar said: (Aug 28, 2012) | |

How do you get 4 members in group? please explain. |

Kashika Shekhar said: (Sep 14, 2012) | |

I am not able to apply different types of logic in different questions. Please tell me some logic for appying the formulas. |

Anusha said: (Jun 30, 2013) | |

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2). Why to change only the first term and last term with formula. Why don't the formula apply to the other terms. Please Explain? |

Imran said: (Aug 24, 2013) | |

See 4C4 means equal to 1 bcoz, there is only one possible way to have all 4 boys i.e 4 out of 4, So, 10C4 - 4C4 now apply on formula N!/n!(N-n)!. 10!/4!(10-4)! -4C4 = 10!/4!(6!)-4C4. = 10*9*8*7*6*5*4*3*2*1/4*3*2*1 * 6*5*4*3*2*1 -4C4. = 10*9*8*7/4*3*2*1 -4C4 (bcoz 6*5*4*3*2*1 cut up and down). = 5040/24 -4C4. = 210 -4C4. = 210 - 1 = 209 answer, (bcoz 4C4 = 1). |

Sam said: (Aug 25, 2013) | |

Is there anyone to explain why this particular question cannot be solved in the same method which were solved above? |

Mohit said: (Aug 26, 2013) | |

Can anyone please help me to reduce the time spent in lengthy calculations. I understood how it can be solved. But calculations take so much time and we don't have much time in solving one particular question. If I try to solve fast, small silly mistake happens and could not get answer in 1st attempt which increases the time spent to re-check it. |

Baba said: (Sep 11, 2013) | |

You could also take a shortcut and say that there are 10*9*8*7 / 4! = 210 ways to arrange 10 people in a group of 4. Then subtract the number of ways to arrange 4 girls in a group of 4 4*3*2*1 / 4! = 1. So 210 Total - 1 All Girl = 209 At least 1 Boy. |

Kiran said: (Sep 13, 2013) | |

There is a trick here just oppose the question like following: We have 6 Boys and 4 Girls, 1) Total No. of Possible ways = 10c4 -->210 (selecting 4 out of 10 Children). 2) Tricky here(Oppose) select no boy at all i.e select 4 children from only girls = 4C4 --> 1. Finally : Subtract 2nd from 1st 210-1 = 209 Answer. |

Nishant Khanorkar said: (Sep 17, 2013) | |

Best way to solve such problem is by going the other way round. At least 1 boy = total - all girls. So, 10c4-4c4 = (10*9*8*7)/(1*2*3*4) - 1. = 210 - 1. = 209. And here it is. out required answer. |

Ravi said: (Dec 23, 2013) | |

Can also be written as: No.of ways (at least one boy) = Total no.of ways - No.of ways with no boy = 10c4 - 4c4 = 210 - 1 = 209. |

Alex said: (Feb 19, 2014) | |

I completely don't get it! Combintations: 1B+3G 2B+2G 3B+1G 4B Those are very clear, no doubts. so, the final answer is (1b+3g)+(3b+2g)+(3b+2g)+4b Let's see the first one. 1b+3g=BGGG (as we have 4 spots). So, for the boy we have 6 different options (as any of the 6 boys can be chosen). We have 3 spots for girls and 4 girls to fill those 3 spots. So, for the first of the 3 girl spots there are 4 different options to choose from (as we have 4 girls), for the second spot we have only 3 girls left (as the forth girl already took the first spot), and for the 3d spot we have 2 girls left. So, the total result for 1b+3g would be 6x4x3x2+ we have to account the other combinations (2b+2g, 3b+1g, 4b). So, how the result is only 209? |

Nikhil said: (Mar 3, 2014) | |

At least 1 boy = total selections - no boy. = 10c4-(6c0*4c4). = 210-1. = 209. |

Prabhu said: (Jul 31, 2014) | |

We may have(1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2). Why to change only the first term and last term with formula. Why don't the formula apply to the other terms. Please Explain? |

Vijay said: (Sep 10, 2014) | |

We may have(1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2). In 1st step : ... + (6c1 x 4c3) +...6C4 In 2nd step : ... + (6c1 x 4c1) +...6C2 How ? Why don't the formula nCr = nC(n-r). So 4C3 = 4C(4-3) = 4C1 apply to the other terms. Please Explain? |

Hitesh said: (Sep 21, 2014) | |

What about when there was asked for at most 2 boys? |

Aditi Muley said: (Nov 4, 2014) | |

In third step last 1 is 6C2, I am confused in the fourth step about how we solved 6C2 that we got 15 in result. |

Shreu said: (Nov 28, 2014) | |

But the also there the children how it is possible? |

Shilpa said: (Dec 3, 2014) | |

6c2= 6!/2!(6-2)! = 6!/2!*4! = 6*5*4! 2*4! Then 4! will be cancel. = 6*5/2. By solving it. = 15 got in result. |

Maggi said: (Dec 19, 2014) | |

In 1st step :........+ (6c1 x 4c3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4). In 2nd step :........+ (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2). If we use formula nC(n-r) then it is not applied to 6C2 x 4C2. So on. |

Mayur said: (Mar 11, 2015) | |

One Question has many times came no one gave answer. Why formula use for only first term ncr(n-r)? Why not for other one, how can we know when we use this formula? |

Srinivas said: (May 6, 2015) | |

Can any one explain how we apply formula ncr = nc(n-r) to only first and last term only? |

Atul said: (Jul 2, 2015) | |

I am not satisfied from the above ways because I don want to use simpler method. Tell me why did we apply formula only on 1st and last terms. Why not on other terms? |

Galla said: (Jul 8, 2015) | |

Wow your various explanations really did help. |

Sarasa said: (Aug 8, 2015) | |

Why didn't nC (n-r) method use for 6c2? Anyone please explained it? I didn't get answer for this. |

Nikhil said: (Aug 16, 2015) | |

Actually one need not use nCr=nC (n-r) concept. Just use the concept that nCr= n!/(r!* (n-r) !). You will get the answer. |

Gunavel said: (Sep 4, 2015) | |

Any another easy method? |

Payal said: (Sep 22, 2015) | |

Second step is not clear. |

Rahul said: (Sep 27, 2015) | |

Why we are not using nc(n-r) formula in all in 2nd step? |

Manek said: (Nov 1, 2015) | |

I have understood this sum. |

Shubham said: (Nov 19, 2015) | |

Guys just calculate the opposite of that, case where no boy is there. 4c4 = 1. Then all case which are (4+6) c4 = 210. Then answer is 210-1 = 209. |

Mujhse said: (Dec 1, 2015) | |

What if the question was at least 2 boys? Need help. |

Sugumar said: (Dec 30, 2015) | |

Why didn't you take 4 girls out of 4 girls? |

Rishi Jain said: (Jan 20, 2016) | |

Hi @Aditya, In 1st step : ......+ (6c1 x 4c3) +...... In 2nd step : ......+ (6c1 x 4c1) +...... How ? You say that, Its because nCr = nC (n-r). So 4C3 = 4C (4-3) = 4C1. Its one of the formula. But I have confused. That what about 6c1 we should also do 6c1 = 6c (6-1) = 6c5. But we didn't know why? Please explain me. |

Kavya said: (Feb 6, 2016) | |

Why there is no 4C4 means 4 girls? |

Amitabha said: (Apr 10, 2016) | |

We have to select 4 children's. So possibilities, as you have explained, are 1 boy, 3 girls, 2 boys, 2 girls, 3 boys, 1 girl and 4 boys. But I think we can consider 4 girls also. |

Himanshu Kulkarni said: (May 20, 2016) | |

The question says at least one boy is to be selected. Selecting 1 boy from 6 boys is 6C1 i.e. 6 ways. And from remaining 9 children 3 are to be selected randomly (any combination of girls and boys) which is 9C3. So why not 6C1 x 9C3 matches the answer? |

Chiru said: (Jul 9, 2016) | |

Its easy. 10c4 - 4c4 =209. |

Sharad said: (Jul 17, 2016) | |

@Udaya. May I know how could you find out the answer from below calculation? So the number of ways the selection can be made such that at least one boy should be there is =10C4 - 4C4 = 209. Please explain this. |

Sonu said: (Aug 25, 2016) | |

Should not we have to arrange all the children? In this we have just make combinations. |

Kayalvizhi said: (Aug 28, 2016) | |

Can anyone explain how to derive this calculation? Please. |

Hmy Nj said: (Sep 6, 2016) | |

Why is it wrong 6C1 * 9C3? |

Alok said: (Sep 29, 2016) | |

Can it not be written as 6 * 9C3? |

Divyae said: (Oct 8, 2016) | |

@Alok and @Hmy Nj We cannot do 6C1 * 9C3 as there will be some repeated selections in this case. Eg, B1, B2, B3, B4, B5, B6 G1, G2, G3, G4 Consider in the first step you selected B1, you have options of selecting any boy in step 2. Now, if you consider selecting B2 in the first step, then you can select B1 in step 2. Thus it creates similar selections. I hope it is clear. :) |

Badri said: (Nov 23, 2016) | |

= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) (6C1 x 4C3) AND (6C4)applied to ncr. |

Daniel said: (Dec 7, 2016) | |

Why not 4 girls? |

Neema said: (Dec 8, 2016) | |

Not mention that how many person should we take at a time, the how it will be solved? |

Shivaraj said: (Dec 18, 2016) | |

@Neema. It's mentioned, read question clearly. |

Pranali said: (Jan 4, 2017) | |

In first step (6c2 * 4c2). Send step ( (6*5/2*1) * (4*3/2*1) ). How? I can't understand it |

Harika said: (Jun 22, 2017) | |

I'm bit confused that why here use multiplication? Please tell me. |

Tarun said: (Jun 26, 2017) | |

@Harika. It's because those two events are needed to complete the work. If any one event is needed to complete the work, we use addition. Hope this helped. |

Shiva said: (Dec 13, 2017) | |

I am confused in the second step, Please explain it. |

Aarav said: (Mar 8, 2018) | |

We can also apply nCr = n!/((r!)(n-r)!). It's one of the formulae. |

Heenu said: (May 31, 2018) | |

Easiest way: Since 4 students are to be selected and there are 4 girls, hence we don't want a combination consisting of all girls: 4c4 =1. Total no of combinations available: (Total students= 6 boys + 4 girls =10 and 4 need to be selected) 10c4= 210. Hence favorable combinations = 210 -1 = 209. |

Sandy said: (Aug 3, 2018) | |

@ALL. So you can select 1 boy in 6C1 ways. So the rest 3 of them can be any gender right. So why can't it be done in 9C3 ways? Why is the final answer not 6C1*9C3? |

Tab said: (Sep 22, 2018) | |

Thanks @Udaya. |

Imnikesh said: (Oct 10, 2018) | |

How 6c4 is reduced to 6c2? |

Aniket said: (Oct 26, 2018) | |

@All. 6c4 = 6c2. Because 6c4= 6!/(4!*2!) and 6c2=6!/(2!*4!) both are same that's why 6c4=6c2. |

Shivam said: (Apr 27, 2019) | |

No boys selected=4c4 ways i.e 1. Selection of 4 from all =10c4 ways=10*9*8*7/4*3*2*1=210 ways. Atleast 1 boy=210-1 = 209 ways. |

Dhanraj said: (Sep 6, 2019) | |

Thanks for explaining it in detail @Shivam. |

Sathvika said: (Sep 25, 2019) | |

4c2 = 4!/2!*(4-2)!. = 4 * 3/2 * 2, = 3. |

P.Krishna Reddy said: (Feb 11, 2020) | |

Atleast one = Total - None = 10c4-1 = 209. |

Nithin said: (Aug 29, 2020) | |

Why can't we do 6C1*9C3? First choose a boy from group of 6 and then choose any three from remaining 9 people. |

Huma_ said: (Jan 11, 2021) | |

Can anyone explain this 2nd step:-. (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4). = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2). Why the nCr formula is applied to only (6C1 x 4C3) & (6C4) and not to all? Please explain. |

Ezhil said: (Mar 11, 2021) | |

@Huma. @Imnikesh. 4c3 becomes 4C1, Let me explain 4C3 means we have to take 3 girls out of 4 girls, generally by using the aforementioned formula was nCr = nC (n-r), this formula is only should be used on the condition that is value is half of the number greater than n value i.e. nCr = 4C3, r=3 value which means are value is half of the value of n=4 increased (half of the value of n=4 is 2 I. E r=3 > 2) so that we applying nCr = nC (n-r) formula, 4C3 becomes 4C1. On the other case 6C1 we should not be applied this formula because r=1 value is smaller the half of the value of n=6 (half of the value of n=6 is 3 i.e r=1 < 3) that's why we could not be used aforesaid nCr formula. |

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