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From $6$ men and $4$ ladies, a committee of $5$ is to be formed. In how many ways can this be done, if the committee is to include at least one lady?









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Option(A) is correct

To committee can be formed in the following ways:

1 lady + 4 gents or 2 ladies + 3 gents or 3 ladies + 2 gents or 4 ladies + 1 gent or 5 ladies + 0 gents.

Total number of possible arrangements:

$({^4C_1} × {^6C_4}) + ({^4C_2} × {^6C_3}) + ({^4C_3} × {^6C_2}) + ({^4C_4} × {^6C_1}) $

⇒ $60+120+60+6$

$= \textbf{246}$

(11) Comment(s)


Can we solve it like this?

$^{10}C_5 - ^6C_5$

it makes sense, right?


No, you can not do this, latter term $^6C_5$ doesn't specify that this includes only men (since you are trying to negate the numbers which originate from selecting ONLY men).

These are ANY six people that may contain ladies as well.


i think all we care about is time and solution to the problem.

Can you please be a bit rational


I tried to pointed out the mistake in your approach. Maybe you didn't like it.

I don't get why have you ask it then?

P.S. NOT everyone is looking for time and solution few of us want to get the concept.


Pretty that's correct actually.. Indirect approach will save time. Finding possible arrangements for one or more women can be seen as ALL possible arrangements - arrangements with NO women


Could someone pls explain why we do not multiply the end solution with 5! ?


Because ORDER doesn't matter here.

For example if you select 2 ladies $(l_1, l_2)$ and three gents $(g_1, g_2,g_3)$, our job ends here.

It does not matter if the list contains $l_1, l_2, g_1, g_2,g_3$ or $l_2, l_1, g_1, g_2,g_3$ or some other combination, since order of the committee selected here is NOT a concern.

Kushal Singh

why can't we do like these

$({^4C_1} \times {^9C_4})$


We can not do so, since doing this will leave many combinations repeated.

Think about it or let me know if you need further explanation.


This questions seems to be flawed. The original scope of the data states that there are ten (10) total individuals to choose from - six (6) men and four (4) ladies.

Yet, in your "correct" answer, you state that one of the committee options could be five (5) ladies and one (1) male.

Where did the extra woman come from?

Ramesh Sidhdhant

Answer does state that, but mathematical expression below that is ignoring it. Had you read the full solution, you would have got it.

P.S. Learn to respect the efforts rather than whining over everything.

You are getting the content for free, there are bound to be some errors, show some politeness, some gratitude man. Huh