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Of the 12 temporary employees in a certain company, 4 will

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chetan2u
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Re: Of the 12 temporary employees in a certain company, 4 will [#permalink] Of the 12 temporary employees in a certain company, 4 will   18 Oct 2020, 06:40
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altairahmad wrote:
chetan2u wrote:
altairahmad wrote:

Hi VeritasKarishma Bunuel chetan2u

Why have we divided by 3! here instead of by 4!/3! ? Isn't the number of ways WWWM can be arranged in 4!/3! ?

Thanks in advance.



You have taken WWWM, but order does not matter while choosing W.
5*4*3 means we have taken order in consideration, that is it is permutation 5P3. To convert this back to combinations where order does not matter, that is 5C3, we require to divide by the way these 3 Ws can be arranged within themselves and that is 3!.

5P3=5!/(5-3)!
5C3=5!/3!(5-3)!.......division by 3! is over and above 5P3.


Thanks for the reply.

Order doesn't matter for the whole WWWM arrangement right ? If we were told to find different arrangements of WWWM it would have been 4!/3! right ? Then why is in this case the arrangements just 3! ? It's as if M can't change the position.



Because, in WWWM you are taking arrangement of just the Ws and 7C1 is the selection of one M, so in 5*4*3*7, the arrangement is for 5*4*3, so divide it by 3!=5*4*3/3!
But you can also select M in 7 ways so multiply by 7=> 7*5*4*3/3!
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Re: Of the 12 temporary employees in a certain company, 4 will [#permalink] Of the 12 temporary employees in a certain company, 4 will   18 Oct 2020, 07:41
Given: Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees.
Asked: If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?

Favorable ways = 5C3 * 7C1 = 10*7 = 70

IMO D
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Of the 12 temporary employees in a certain company, 4 will [#permalink] Of the 12 temporary employees in a certain company, 4 will   18 Oct 2020, 08:29
chetan2u wrote:


You have taken WWWM, but order does not matter while choosing W.
5*4*3 means we have taken order in consideration, that is it is permutation 5P3. To convert this back to combinations where order does not matter, that is 5C3, we require to divide by the way these 3 Ws can be arranged within themselves and that is 3!.

5P3=5!/(5-3)!
5C3=5!/3!(5-3)!.......division by 3! is over and above 5P3.


Thanks for the reply.

Order doesn't matter for the whole WWWM arrangement right ? If we were told to find different arrangements of WWWM it would have been 4!/3! right ? Then why is in this case the arrangements just 3! ? It's as if M can't change the position.[/quote]


Because, in WWWM you are taking arrangement of just the Ws and 7C1 is the selection of one M, so in 5*4*3*7, the arrangement is for 5*4*3, so divide it by 3!=5*4*3/3!
But you can also select M in 7 ways so multiply by 7=> 7*5*4*3/3![/quote]

Thanks again for the reply.

Actually I am still not sure I understand. In what case would it have been ok to divide by 4!/3! ? Maybe by comparing the two scenarios I can get the crux of the issue.

What I have usually been doing is this : Find the number of ways WWWM can be arranged (order matters scenario) and then divide by the number of arrangements to get the unique combinations in which order does not matter i.e 4!/3!. It was working so well until now.....😭😭

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Re: Of the 12 temporary employees in a certain company, 4 will [#permalink] Of the 12 temporary employees in a certain company, 4 will   18 Oct 2020, 18:54
chetan2u wrote:
altairahmad wrote:
chetan2u wrote:

Because, in WWWM you are taking arrangement of just the Ws and 7C1 is the selection of one M, so in 5*4*3*7, the arrangement is for 5*4*3, so divide it by 3!=5*4*3/3!
But you can also select M in 7 ways so multiply by 7=> 7*5*4*3/3!


Thanks again for the reply.

Actually I am still not sure I understand. In what case would it have been ok to divide by 4!/3! ? Maybe by comparing the two scenarios I can get the crux of the issue.

What I have usually been doing is this : Find the number of ways WWWM can be arranged (order matters scenario) and then divide by the number of arrangements to get the unique combinations in which order does not matter i.e 4!/3!. It was working so well until now.....😭😭
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Re: Of the 12 temporary employees in a certain company, 4 will [#permalink] Of the 12 temporary employees in a certain company, 4 will   27 Nov 2020, 00:17
Hi @Bunuel,@chetan2u,
This one although it is very easy to solve tricked me a lot with the wording. So in my understanding i can not find out where am I faltering:
Since 4 employees are about to leave and we want to find the possible number of teams consisting of 3W and 1M then the following strategy should be applied.
For the question to hold true number of women leaving should not be greater than 2 in order to at least have 3 women in the group ( women in the beginning - max possible number of leaves).
So the cases are as follows: leaving 2W and 2M ,leaving 1W and 3M, leaving 0W and 4M. This results into 3C3*5C1+4C3*4C1+5C3*3C1=51.
Can you please enlighten me?
TIA
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Re: Of the 12 temporary employees in a certain company, 4 will [#permalink] Of the 12 temporary employees in a certain company, 4 will   27 Nov 2020, 00:57
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A very simple question on combinations (choosing)

For those who have a doubt: Permutations means arrangement and combinations means choosing.

The question clearly states that it will be employing 4 people from 12 employees of which 5 are women and the remaining 7 are men.

It does not state anything about 4 employees leaving.


Out of the 4 people being chosen, 3 have to be women and 1 has to be a man. There is no arrangement here.

So out of 5 women choose 3 = 5C3 = \(\frac{5!}{(5 - 3)!3!} = \frac{5!}{2!3!} = 10\)

Out of 7 men choose 1 = 7C1 = \(\frac{7!}{(7 - 1)!1!} = \frac{7!}{6!1!} = 7\)

Since it is 3 women and 1 man, we multiply = 10 * 7 = 70


Option D

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Re: Of the 12 temporary employees in a certain company, 4 will [#permalink] Of the 12 temporary employees in a certain company, 4 will   25 Jan 2021, 13:59
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themagiccarpet wrote:
Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?

A. 22
B. 35
C. 56
D. 70
E. 105


3 women member and 1 man member

\(5^c_3*7^c_1=70\)

The answer is \(D\)
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Re: Of the 12 temporary employees in a certain company, 4 will [#permalink] Of the 12 temporary employees in a certain company, 4 will   27 Jan 2021, 18:53
(5 x 4 x 3) / 3! = 60/6 = 10 <---- # of ways women can be selected for a group consisting of three women and 1 man

7 <--- # of ways a man can be selected for that 1 position

10 x 7 = 70

Ans is D. IMO
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Re: Of the 12 temporary employees in a certain company, 4 will [#permalink] Of the 12 temporary employees in a certain company, 4 will   02 May 2024, 11:36
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Re: Of the 12 temporary employees in a certain company, 4 will [#permalink]
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