### Discussion :: Permutation and Combination - General Questions (Q.No.11)

Priya said: (Jun 26, 2010) | |

How? I can't understand this problem. |

Sahithya said: (Sep 10, 2010) | |

Hi priya,. We have to select 5men out of 7 men, this can be done in 7C5 ways = 7C2 [Ref: nCr=nC (n-r) ]. And also we hav to select 2women out of 3 women, this can be done in 3C2 ways = 3C1. As these two combinations are compulsry, we hav to multiply 7C2 and 3C1. I think you understand. |

Binushree said: (Nov 25, 2010) | |

How to multiply? |

Karthik said: (Jan 7, 2011) | |

Why can't we multiply 7C5 nd 3C2 directly ? |

Suvarna said: (Feb 25, 2011) | |

It takes more time to calculate 7c5 than 7c2& 3C2 than 3C1, to improve the speed we are doing like this. |

Jat said: (Jun 4, 2011) | |

(7C5 x 3C2) = (7C2 x 3C1) This step i can't understand. Please explain me. |

Manasa said: (Jun 14, 2011) | |

@jat its just formula n =n c c r n-r n=7,r=5 7C5=7C2 |

Zalak said: (Jul 30, 2011) | |

(7C5 x 3C2) = (7C2 x 3C1) This step I can't understand. Please explain me. |

Kavu said: (Sep 18, 2011) | |

(7C2 x 3C1) = (7/2 x 6/1) x 3= 63. How it comes? |

Achutharaj said: (Sep 28, 2011) | |

Basic formula nCr = nCn-r so 7C7-5 = 7C2 3C3-2 = 3C1 |

Divya said: (Sep 29, 2011) | |

How 3c2 change into 3c1? |

Ankur Anand said: (Oct 8, 2011) | |

Why combination why not permutation ? Similar as that of the arranging of letters the people are diffrent so there can be many ways by permutation. |

Nirmal said: (Nov 25, 2011) | |

(7C2 x 3C1) = 7 x 6 x 3 How it happened explain please? |

Teja said: (Dec 30, 2011) | |

Hey see you said that 7c5*3c2 is long procedure. That the thing we take ncr formula. ?and second thing why we take that formula in this problem? am not? |

Srikanth said: (Apr 26, 2012) | |

@teja. As this a problem f selection you use combination. It means calculating 7c5 is a bit more tedious when compared to calculating 7c2. |

Moni said: (Jun 27, 2012) | |

For ex. Take 7C5, 5 is more than a half of 7. So we can use the formula nC (n-r) when are is more than half of n. Similarly 3C2 is made as 3C1. I hope you have understood. |

Niyati Bafna said: (Jul 27, 2013) | |

Are there any more problems, perhaps of a slightly advanced kind, on the net? If so, on which sites can they be found? |

Vinod Rathod said: (Sep 16, 2013) | |

Lets do by this way : As you said that (7c5*3c2) = (7c2*3c1). 7/5*3/2 = 7/2*3/1 = 63. |

Pooja said: (Sep 28, 2013) | |

I understood that thanks! we have to use nCr=nCn-r. |

Chandan said: (Oct 30, 2013) | |

Why multiplication not addition? |

Aswin said: (Apr 2, 2014) | |

Another question : how to select 6 people from a group of 10 so that two of them does not come together? |

Satish said: (May 10, 2014) | |

Step 1: 7C5 = (7*6*5*4*3)/(5*4*3*2*1) [ 7C5=> here from 7 factorial for 5 counts and divide totally by 5 factorial]. = (7*6)/(2*1) [here group the combinations again]. = 7C2. So 7C5=7C2 = (7*6)/2 =21 ---->1. Step 2: similarly 3C2= (3*2)/(2*1) = 3/1= 3 ----> 2. Multiply(Combination) (1) and (2) = Ans = 63. [ ie.. nCr = n!/r! = nC(n-r)]. |

Shital said: (Dec 9, 2014) | |

I didn't understand 7c2*3c1. |

Vishesh said: (Dec 9, 2014) | |

Same here what is this c? I don't understand this. |

Manju said: (May 24, 2015) | |

nCr = n!/(n-r)!r! = 7C5*3C2. = (7*6/2*1)*3 = 63. |

Bond81 said: (Jun 23, 2015) | |

According to me the answer should be 126 as: If we consider men in 1 group that is 7c5 and then women in 1 group that is 3c2 we can arrange these two groups in 2! ways. So answer should be 7c5*3c2*2! = 126. |

Bgt said: (Aug 8, 2015) | |

How can't we use any other e example? |

Sravani said: (Sep 24, 2015) | |

How 3c2 = 3c1? |

Harish Julai said: (Oct 7, 2015) | |

Hello @Sravani. If n>r, then ncr becomes nc(n-r). |

Kunal said: (Nov 3, 2015) | |

But why they did 7c5*3c2 as 7c2*3c1? Why they directly not calculate 7c5*3c1? |

Shivam Shukla said: (Dec 1, 2015) | |

I didn't understand this problem. Please each and every step explain it. |

Anoushka said: (Feb 26, 2016) | |

Sir I haven't understood why have we multiplied 3 with 7*6/2*1 *3. |

Sam said: (Jun 18, 2016) | |

Why we are taking as 7c2 * 3c1? Please explain it. |

Sudha said: (Sep 7, 2016) | |

How 7c2 will come instead 7C5? |

Mayank said: (Sep 23, 2016) | |

@Sudha. As know this formula n!/r!(n-r)!. 7!/5!2! 3!/2!1! The answer will be same . What they have done is try to reduce one step. |

Mariyam said: (Oct 13, 2016) | |

@Satish your explanation is very understandable. |

Dhayithri said: (Nov 11, 2016) | |

Here we can also taken as 3c2 as 3c1 because factorial of those two is equal. |

Feze said: (Dec 7, 2016) | |

Why we multiply with 3 here? |

Jyoti said: (Dec 15, 2016) | |

According to me, the final answer which displayed is wrong. If we multiply 126 is coming. |

Shivaraj said: (Dec 18, 2016) | |

@Feze. 3c1=3 By std formulae. |

Naresh said: (Dec 20, 2016) | |

How to get the answer? Please anybody explain me. |

Thatchayini said: (Jan 11, 2017) | |

In that question, they didn't mention anything about permutation or combination. so we can try it in both ways. 7C5 * 3C2 = (7 * 6 * 5 * 4 * 3)/(5 * 4 * 3 * 2 * 1) * (3 * 2)/(2 * 1). = 7 * 3 * 3 = 21 * 3 = 63. |

Kyosi Billy said: (Aug 26, 2017) | |

Thanks for explaining it. |

Bhavani said: (Mar 2, 2018) | |

How will get 7c2* 3c1? Please explain me. |

Shree said: (May 21, 2018) | |

Thanks for giving an explanation. |

Mike said: (Jul 9, 2018) | |

Where 6 comes from? Please explain me. |

Ashutosh Sahoo said: (Aug 25, 2018) | |

5 out of 7 and 2 out of 3. so, =7c5 x 3c2. = (7! / 5! x 2!) x (3! / 2! x 1!), = (7 x 6 / 2) x 3, =( 42 / 2 ) x 3, = 21 x 3, = 63 (ans). |

Keerthana said: (Sep 11, 2018) | |

The answer is 63. I agree. |

Kamal said: (Sep 14, 2018) | |

Anybody help to solve this problem? 7 members have to be selected from 12 men and 3 women such that no two women can come together in how many ways we can select them? |

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