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Combinatorics problems:

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by \'manpreet singh » Tue Jan 15, 2013 8:59 pm
Kevin has wired 6 light bulbs to a board so that, when he presses a button, each bulb has
an equal chance of lighting up or staying dark. Each of the six bulbs is independent of the
other five.
(a) In how many different configurations could the bulbs on the board light up (including
the configuration in which none of them light up)?
(b) In how many configurations could exactly three of the bulbs light up?
(c) When Kevin presses the button, what is the probability that exactly three of the
bulbs light up?
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by GMATGuruNY » Wed Jan 16, 2013 4:37 am
'manpreet singh wrote:Kevin has wired 6 light bulbs to a board so that, when he presses a button, each bulb has
an equal chance of lighting up or staying dark. Each of the six bulbs is independent of the
other five.
(a) In how many different configurations could the bulbs on the board light up (including
the configuration in which none of them light up)?
Number of options for the first bulb = 2. (On or off.)
Number of options for the second bulb = 2. (On or off.)
Number of options for the third bulb = 2. (On or off.)
Number of options for the fourth bulb = 2. (On or off.)
Number of options for the fifth bulb = 2. (On or off.)
Number of options for the sixth bulb = 2. (On or off.)
To combine these options, we multiply:
2*2*2*2*2*2 = 64.
(b) In how many configurations could exactly three of the bulbs light up?
Number of options for the first illuminated bulb = 6. (Any of the 6 bulbs.)
Number of options for the second illuminated bulb = 5. (Any of the 5 remaining bulbs.)
Number of options for the third illuminated bulb = 4. (Any of the 4 remaining bulbs.).
To combine these options, we multiply:
6*5*4.
Since the ORDER of the bulbs doesn't matter -- ABC and CAB represent the same configuration of illuminated bulbs -- we divide by the number of ways to ARRANGE the 3 bulbs (3!):
(6*5*4)/(3*2*1) = 20.
(c) When Kevin presses the button, what is the probability that exactly three of the
bulbs light up?
As determined above:
Number of configurations of 3 that could light up = 20.
Total number of ways that the 6 bulbs could light up = 64.
Thus:
(Configurations of 3)/(Total ways) = 20/64 = 5/16.
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by vishugogo » Thu Jan 17, 2013 1:07 am
GMATGuruNY...just in case in the second question...the question had asked for

In how many configurations could atlease three of the bulbs light up?

how would we solve it...

GMATGuruNY wrote:
'manpreet singh wrote:Kevin has wired 6 light bulbs to a board so that, when he presses a button, each bulb has
an equal chance of lighting up or staying dark. Each of the six bulbs is independent of the
other five.
(a) In how many different configurations could the bulbs on the board light up (including
the configuration in which none of them light up)?
Number of options for the first bulb = 2. (On or off.)
Number of options for the second bulb = 2. (On or off.)
Number of options for the third bulb = 2. (On or off.)
Number of options for the fourth bulb = 2. (On or off.)
Number of options for the fifth bulb = 2. (On or off.)
Number of options for the sixth bulb = 2. (On or off.)
To combine these options, we multiply:
2*2*2*2*2*2 = 64.
(b) In how many configurations could exactly three of the bulbs light up?
Number of options for the first illuminated bulb = 6. (Any of the 6 bulbs.)
Number of options for the second illuminated bulb = 5. (Any of the 5 remaining bulbs.)
Number of options for the third illuminated bulb = 4. (Any of the 4 remaining bulbs.).
To combine these options, we multiply:
6*5*4.
Since the ORDER of the bulbs doesn't matter -- ABC and CAB represent the same configuration of illuminated bulbs -- we divide by the number of ways to ARRANGE the 3 bulbs (3!):
(6*5*4)/(3*2*1) = 20.
(c) When Kevin presses the button, what is the probability that exactly three of the
bulbs light up?
As determined above:
Number of configurations of 3 that could light up = 20.
Total number of ways that the 6 bulbs could light up = 64.
Thus:
(Configurations of 3)/(Total ways) = 20/64 = 5/16.
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by vishugogo » Sat Jan 19, 2013 7:13 pm
GMATGuruNY...help please
vishugogo wrote:GMATGuruNY...just in case in the second question...the question had asked for

In how many configurations could atlease three of the bulbs light up?

how would we solve it...

GMATGuruNY wrote:
'manpreet singh wrote:Kevin has wired 6 light bulbs to a board so that, when he presses a button, each bulb has
an equal chance of lighting up or staying dark. Each of the six bulbs is independent of the
other five.
(a) In how many different configurations could the bulbs on the board light up (including
the configuration in which none of them light up)?
Number of options for the first bulb = 2. (On or off.)
Number of options for the second bulb = 2. (On or off.)
Number of options for the third bulb = 2. (On or off.)
Number of options for the fourth bulb = 2. (On or off.)
Number of options for the fifth bulb = 2. (On or off.)
Number of options for the sixth bulb = 2. (On or off.)
To combine these options, we multiply:
2*2*2*2*2*2 = 64.
(b) In how many configurations could exactly three of the bulbs light up?
Number of options for the first illuminated bulb = 6. (Any of the 6 bulbs.)
Number of options for the second illuminated bulb = 5. (Any of the 5 remaining bulbs.)
Number of options for the third illuminated bulb = 4. (Any of the 4 remaining bulbs.).
To combine these options, we multiply:
6*5*4.
Since the ORDER of the bulbs doesn't matter -- ABC and CAB represent the same configuration of illuminated bulbs -- we divide by the number of ways to ARRANGE the 3 bulbs (3!):
(6*5*4)/(3*2*1) = 20.
(c) When Kevin presses the button, what is the probability that exactly three of the
bulbs light up?
As determined above:
Number of configurations of 3 that could light up = 20.
Total number of ways that the 6 bulbs could light up = 64.
Thus:
(Configurations of 3)/(Total ways) = 20/64 = 5/16.
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by GMATGuruNY » Sat Jan 19, 2013 8:15 pm
vishugogo wrote:GMATGuruNY...just in case in the second question...the question had asked for

In how many configurations could atlease three of the bulbs light up?

how would we solve it...
In my solution above, here is how we counted the number of ways 3 bulbs could light up:
Number of options for the first illuminated bulb = 6. (Any of the 6 bulbs.)
Number of options for the second illuminated bulb = 5. (Any of the 5 remaining bulbs.)
Number of options for the third illuminated bulb = 4. (Any of the 4 remaining bulbs.).
To combine these options, we multiply:
6*5*4.
Since the ORDER of the bulbs doesn't matter -- ABC and CAB represent the same configuration of illuminated bulbs -- we divide by the number of ways to ARRANGE the 3 bulbs (3!):
(6*5*4)/(3*2*1) = 20.
Following this same reasoning:
Number of ways for 4 bulbs to light up = (6*5*4*3)/(4*3*2*1) = 15.
Number of ways for 5 bulbs to light up = (6*5*4*3*2)/(5*4*3*2*1) = 6.
Number of ways for 6 bulbs to light up = (6*5*4*3*2*1)/(6*5*4*3*2*1) = 1.

Thus:
Number of ways for AT LEAST 3 bulbs to light up = 20+15+6+1 = 42.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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by \'manpreet singh » Mon Jan 21, 2013 10:26 pm
Thanks for the clear explanation as always. :)

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Singh
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