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by Veronica » Sun Sep 26, 2010 8:09 pm
1. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
a. 20, b.40, c.50, d.80, e. 120

2. A certain business produced x rakes each month form Nov through Feb and shipped x/2 rakes at the beginning of each month from March through Oct. The business paid no storage costs for the rakes from Nov through Feb, but it paid storage costs of $0.1 per rake each month from March through October for the rakes that had not been shipped. In terms of x, what was the total storage cost, in dollars, that the business paid for the rakes for the 12 months form Nov through Oct?
a. 0.4x, b.1.2x, c.1.4x, d.1.6x, e.3.2x

Please help me solve them in shortest time, thanks!
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by Rahul@gurome » Sun Sep 26, 2010 8:35 pm
(1) Solution:
There are 5 married couples. Or there are 2*5 = 10 people in total.
Number of ways of selecting 3 people from among 10 is 10C3 = 120 ways.

Let us find the number of ways in which we can select 3 people in a way that 1 couple is always there.
One couple can be selected from among the five in 5C1 or 5 ways.
So the third person can be selected from (10 - 2) or 8 people in 8C1 or 8 ways.
Or the number of ways in which we can select 3 people in a way that 1 couple is always there is 5*8 = 40.

Hence the number of ways of selecting 3 people such that there is never a couple is 120 - 40 = 80.
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by diebeatsthegmat » Sun Sep 26, 2010 10:50 pm
Veronica wrote:1. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
a. 20, b.40, c.50, d.80, e. 120

2. A certain business produced x rakes each month form Nov through Feb and shipped x/2 rakes at the beginning of each month from March through Oct. The business paid no storage costs for the rakes from Nov through Feb, but it paid storage costs of $0.1 per rake each month from March through October for the rakes that had not been shipped. In terms of x, what was the total storage cost, in dollars, that the business paid for the rakes for the 12 months form Nov through Oct?
a. 0.4x, b.1.2x, c.1.4x, d.1.6x, e.3.2x

Please help me solve them in shortest time, thanks!
for the second question: each month it produced x raked each month from 11 till 2 of year it means it produced 4x
each month from march to oct it sold x/2 and must keep the last products left in somewhere with the fee : 0,1$
march: 4x-x/2=7x/2 and keep 7x/2 with the same price 0.1*6x/2
aprilL 7x/2-x/2=6x/2 and keep 6x/2 with the same price 0.1*6x/2
you do the same till oct... then add em all together
(7/2+6/2+5/2+4/2+3/2+2/2+1/2)x*0.1=1.4x
the correct answer is C
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by Veronica » Sun Sep 26, 2010 11:22 pm
diebeatsthegmat wrote:
Veronica wrote:1. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
a. 20, b.40, c.50, d.80, e. 120

2. A certain business produced x rakes each month form Nov through Feb and shipped x/2 rakes at the beginning of each month from March through Oct. The business paid no storage costs for the rakes from Nov through Feb, but it paid storage costs of $0.1 per rake each month from March through October for the rakes that had not been shipped. In terms of x, what was the total storage cost, in dollars, that the business paid for the rakes for the 12 months form Nov through Oct?
a. 0.4x, b.1.2x, c.1.4x, d.1.6x, e.3.2x

Please help me solve them in shortest time, thanks!
for the second question: each month it produced x raked each month from 11 till 2 of year it means it produced 4x
each month from march to oct it sold x/2 and must keep the last products left in somewhere with the fee : 0,1$
march: 4x-x/2=7x/2 and keep 7x/2 with the same price 0.1*6x/2
aprilL 7x/2-x/2=6x/2 and keep 6x/2 with the same price 0.1*6x/2
you do the same till oct... then add em all together
(7/2+6/2+5/2+4/2+3/2+2/2+1/2)x*0.1=1.4x
the correct answer is C

Can you explain why the price is 0.1*6x/2, please???
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by euro » Sun Sep 26, 2010 11:48 pm
Total rakes produced = 4x

x/2 rakes get shipped in the beginning of every month.
During the first month, 4x-x/2 rakes are to stored.
Hence, storage for
1st month = (4x-x/2) = 7x/2
2nd month = (4x- 2x/2) = 4x-x = 3x
3rd month = 4x-3x/2 = 5x/2
4th month = 4x-4x/2 = 4x-2x = 2x
....
and so on.

Rakes have to be stored for seven months and the total rakes stored =
summation of 7x/2+3x+5x/2+2x+... = 14x
Storage cost = $0.1 * 14x = $1.4x
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by lokesh r » Mon Sep 27, 2010 2:28 am
Rahul@gurome wrote:(1) Solution:
There are 5 married couples. Or there are 2*5 = 10 people in total.
Number of ways of selecting 3 people from among 10 is 10C3 = 120 ways.

Let us find the number of ways in which we can select 3 people in a way that 1 couple is always there.
One couple can be selected from among the five in 5C1 or 5 ways.
So the third person can be selected from (10 - 2) or 8 people in 8C1 or 8 ways.
Or the number of ways in which we can select 3 people in a way that 1 couple is always there is 5*8 = 40.

Hence the number of ways of selecting 3 people such that there is never a couple is 120 - 40 = 80.
Hi,

Can u tell me what is the mistake i am doing in below calculation.

Total no of ppl =10.

Case 1 : Total no of ways of selecting first person=10.
Case 2: Total no of ways of selecting second person =8 (not 9, since we should not select a person who is married to person selected in first case)
Case 3: Total no of ways of selecting third person= 6(not 7,since we should not select a person who is married to person selected in first case and second case )

Total possibilities = 10x8x6=480.
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by euro » Mon Sep 27, 2010 7:54 am
lokesh r wrote: Hi,

Can u tell me what is the mistake i am doing in below calculation.

Total no of ppl =10.

Case 1 : Total no of ways of selecting first person=10.
Case 2: Total no of ways of selecting second person =8 (not 9, since we should not select a person who is married to person selected in first case)
Case 3: Total no of ways of selecting third person= 6(not 7,since we should not select a person who is married to person selected in first case and second case )

Total possibilities = 10x8x6=480.
Hi Lokesh!
Initially, I had done exactly the way you have posted above just to realize that something's wrong!!

What you have done - 10x8x6=480 is permutation. This would hold good if say, we were to find the no. of ways in which the three selected people have to be seated on three distinct chairs or given three prizes 1st, 2nd and 3rd.

Note that 480 = 80 * 3!
There are 3! ways of giving 1st, 2nd and 3rd prizes to the three chosen people. For this problem, we need to use combination since the order of selection does not matter.

Hope this makes it clear :)
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by lokesh r » Mon Sep 27, 2010 1:09 pm
euro wrote:
lokesh r wrote: Hi,

Can u tell me what is the mistake i am doing in below calculation.

Total no of ppl =10.

Case 1 : Total no of ways of selecting first person=10.
Case 2: Total no of ways of selecting second person =8 (not 9, since we should not select a person who is married to person selected in first case)
Case 3: Total no of ways of selecting third person= 6(not 7,since we should not select a person who is married to person selected in first case and second case )

Total possibilities = 10x8x6=480.
Hi Lokesh!
Initially, I had done exactly the way you have posted above just to realize that something's wrong!!

What you have done - 10x8x6=480 is permutation. This would hold good if say, we were to find the no. of ways in which the three selected people have to be seated on three distinct chairs or given three prizes 1st, 2nd and 3rd.

Note that 480 = 80 * 3!
There are 3! ways of giving 1st, 2nd and 3rd prizes to the three chosen people. For this problem, we need to use combination since the order of selection does not matter.

Hope this makes it clear :)

Oh my god!!!!!!
This is combinations problem..i did not realize it..whole day i kept scratchin my head over this problem..thinking why the hell am i getting alien answer..

Thanks buddy..
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by diebeatsthegmat » Mon Sep 27, 2010 5:30 pm
Veronica wrote:
diebeatsthegmat wrote:
Veronica wrote:1. If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
a. 20, b.40, c.50, d.80, e. 120

2. A certain business produced x rakes each month form Nov through Feb and shipped x/2 rakes at the beginning of each month from March through Oct. The business paid no storage costs for the rakes from Nov through Feb, but it paid storage costs of $0.1 per rake each month from March through October for the rakes that had not been shipped. In terms of x, what was the total storage cost, in dollars, that the business paid for the rakes for the 12 months form Nov through Oct?
a. 0.4x, b.1.2x, c.1.4x, d.1.6x, e.3.2x

Please help me solve them in shortest time, thanks!
for the second question: each month it produced x raked each month from 11 till 2 of year it means it produced 4x
each month from march to oct it sold x/2 and must keep the last products left in somewhere with the fee : 0,1$
march: 4x-x/2=7x/2 and keep 7x/2 with the same price 0.1*6x/2
aprilL 7x/2-x/2=6x/2 and keep 6x/2 with the same price 0.1*6x/2
you do the same till oct... then add em all together
(7/2+6/2+5/2+4/2+3/2+2/2+1/2)x*0.1=1.4x
the correct answer is C

Can you explain why the price is 0.1*6x/2, please???
you left 6x/2 from selling x/2 prodduct in a month from 7x/2 thus all you have left is 6x/2 and you have to store 6x/2 products and its fee is $0.1 for any product you stored thus with 6x/2 product you stored you will have to pay 6x/2*0.1

do you get it now?
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