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Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 1)
Permutation and Combination
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?
564
645
735
756
None of these
Answer: Option
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.

Discussion:
137 comments Page 7 of 14.
Ajaykumar said:   1 decade ago
7C3 = 35, 7!/(7-3)!3! = 7!/4!3! = 7*6*5/3*2*1 = 35.
Ajay said:   1 decade ago
7C5 written as 7C2 because 7C5 = 7C(7-5) both having same values from formula nCr = nC(nr).
Maryann said:   1 decade ago
Can someone explain better for me because I'm getting confused?
Naveen Kumar said:   1 decade ago
Why we are calculating 7C5 as 7C2?
Sam said:   1 decade ago
Can you please explain 7c5 is converted into nc(n-r)?

But why didn't convert 6c2 into 6c(6-2)?
Naji said:   1 decade ago
How can we notice that we just have to select without arrangement ?
Ananya said:   1 decade ago
Is there any shortcut for this?
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?
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?
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.
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