In a room there are eight lights. Each light can be switched on and off independently of the others. In how many ways can the room be lit with-
five lights on?
at least five lights on?
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1Do you know how to find the number of ways to choose $5$ lights from $8$ lights? – André Nicolas Dec 29 '14 at 05:25
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i have figured out '5 lights on' as 56,but i can't get at least five lights on.. – Richardparker800 Dec 29 '14 at 05:26
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yes,its 87654 divided by 543*2 which is 56. – Richardparker800 Dec 29 '14 at 05:27
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I know binomial coefficients but i cant get the at least five lights on part.. – Richardparker800 Dec 29 '14 at 05:29
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at least 5 lights means:either (5 OR 6 OR 7 OR 8). Replace "OR" by "+" – Vikram Dec 29 '14 at 05:39
3 Answers
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With respect to the Multiplication Principle
If one desires to light a room with exactly 5 lights - obviously 5 lights must be on out of the possible 8 lights. If one decides to use the multiplication principle, they must adjust the answer to account for the fact that the order in which these lights are on does not matter! Therefore our answer is
$$\mathrm{Exactly\ 5\ Lights}=\frac{8 \times 7 \times 6 \times 5 \times 4.}{5 \times 4 \times 3\times 2\times 1} = \frac{8 \times 7 \times 6 \times 5 \times 4.}{5!}$$
Answering the outcome of at least 5 lights is equivalent to finding the number of outcomes for exactly 5, 6, 7 and 8 lights independently and summing the individual answers.
$$\mathrm{At\ Least\ 5\ Lights}=\frac{8 \times 7 \times 6 \times 5 \times 4}{5!} + \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3}{6!} + \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}{7!} + \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{8!} $$
Of course, these calculations can look nicer using binomial coefficients. If you have not learn't this then this answer is for you.
Gustavo Louis G. Montańo
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To turn exactly 5 lights on we will have to choose any 5 of the 8 available lights, this can be done in $8\choose5$$=56$ ways.
In order to turn at least 5 lights on, we will have to subtract the cases of turning 0,1,2,3,4 lights on from the total number of cases. i.e.
$2^8$-$8\choose4 $-$8\choose3 $-$8\choose2 $-$8\choose1 $-$8\choose0 $$=93$
Dheeraj Kumar
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Why not just do $8\choose 5$+$8\choose 6$+$8\choose 7$+$8\choose 8$? – John Dvorak Dec 29 '14 at 05:30
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@JanDvorak you are right! but, subtracting non required from total looks more intuitive. – Dheeraj Kumar Dec 29 '14 at 05:31
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The number of ways to have five lights on is the number of way to choose (see André Nicolas's comment) a $5$-element subset of a set of $8$ elements. This is what is called a binomial coefficient:
$$\binom{8}{5}$$
For at least five, you can add up all of the possibilities for the number of lights on:
$$\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}$$
Zubin Mukerjee
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