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
Answer: Option D
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
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:
(Thu, Jul 15, 2010 02:25:53 AM)
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:
(Fri, Dec 24, 2010 11:25:33 AM)
Nice answer nagu.
Nikhil said:
(Mon, Dec 27, 2010 02:47:13 AM)
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:
(Sat, Jan 15, 2011 05:20:09 AM)
@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:
(Mon, Mar 14, 2011 09:56:18 AM)
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:
(Sat, May 14, 2011 08:17:26 AM)
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:
(Sun, May 29, 2011 02:10:05 PM)
Not a good answer to give please another method please easy method.
Bhargav said:
(Sun, May 29, 2011 02:12:45 PM)
How can you take a 3 men 2 women 4men and 1women and a 5 men please solve easy method.
Rajeev said:
(Fri, Jul 22, 2011 02:15:28 PM)
In the first step 7c5 converted 7c2 in the second step. But how ?
Renuka said:
(Thu, Aug 11, 2011 12:33:32 AM)
Because nCr = nCn-r.
Kumamako said:
(Tue, Aug 16, 2011 12:24:59 AM)
I think that it should be 3*2^3*3-3
Lana said:
(Sat, Aug 27, 2011 03:18:51 PM)
Hey can any one explain me how 210 came because I got 140 instead of 210.
Lookin forward for your support.
Osei-Tutu said:
(Tue, Sep 20, 2011 01:37:59 PM)
Please what is mean by 7C3. Please help.
Kranthi said:
(Wed, Oct 12, 2011 01:29:35 AM)
How 7C4 became 7C3 ?
Siva said:
(Sun, Oct 16, 2011 12:32:53 PM)
7c4 = 7c(7-3)
So came 7c3 @kranthi...
Based on the formula nCr = nC(n-r).
Leo said:
(Tue, Oct 18, 2011 09:37:09 PM)
--> (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
Shouldn't it be like this? --> (7C3 x 6C2) x (7C4 x 6C1) x (7C5)
Vivacity said:
(Fri, Oct 21, 2011 08:48:45 PM)
@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:
(Wed, Nov 16, 2011 03:19:05 PM)
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:
(Tue, Dec 20, 2011 12:04:07 PM)
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:
(Mon, Dec 26, 2011 03:37:11 PM)
How do you find weather give problem can solved by combinations or permutation?
Anshul Vyas said:
(Tue, Jan 24, 2012 03:03:33 PM)
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.
Required number of ways
