Aptitude - Permutation and Combination - Discussion
Discussion Forum : 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?
Answer: Option
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
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= (7C3 x 6C2) + (7C4 x 6C1) + (7C5) | |||||||||||
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= (525 + 210 + 21) | ||||||||||||
= 756. |
Discussion:
137 comments Page 7 of 14.
VAMSI REDDY 1100 said:
1 decade ago
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.
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:
1 decade ago
It is so long procedure, is any shortcut method for this?
Kanaga said:
1 decade ago
Which formula using this method clearly?
Ankit said:
1 decade ago
@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 :).
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:
1 decade ago
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.
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:
1 decade ago
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.
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:
1 decade ago
= (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)?
= 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:
1 decade ago
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.
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:
1 decade ago
I have a small doubt. Why we didn't divide it like this 7c3*6c2+7c4*6c1+7c5*6c0/13c5?
Mayur said:
1 decade ago
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?
Why didn't we use in 6c1? Can anybody clear me please?
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