Adriana´s room contains N identical light bulbs with independent switc

Question

GMATH  practice exercise (Quant Class 18)

Adriana´s room contains N identical light bulbs with independent switches, and this room is considered well lighted only if at least two of the light bulbs are switched on. If there are exactly 26 different ways to make Adriana´s room well lighted, what is the value of N?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

kunalcvrce · Accepted Answer

xC2+xC3+..+xCx=26 
putting value 5
 5C2+5C3+5C4+5C5=26 
B

KanishkM · Answer

keyword: at least two of the light bulbs are switched on

We are given the total ways, we need to find the n =>  nC_x  = 26

xC_2 + xC_3 + xC_4  = 26

We can try back solving here, Only 5 works perfectly.

5C_2  + 5C_3  +  5C_4  +  5C_5  = 26

B

fskilnik · Answer

Our "official solution" (similar to the previous contributions) is the following:

?\,\,\,:\,\,\,N \ge 4\,\,{\mathop{\rm int}} \,\,{\rm{such}}\,\,{\rm{that}}\,\,\,B\left( N \right) = 26\,,\,\,\,{\rm{where}}\,\,\,B\left( N \right) = C\left( {N,2} \right) + C\left( {N,3} \right) + \ldots + C\left( {N,N} \right)\,\,\,\,\,\,\,\left( * \right)

\left( {\rm{A}} \right)\,\,\,N = 4\,\,\,\,\, \Rightarrow \,\,\,\,B\left( 4 \right)\,\, = \,\,C\left( {4,2} \right) + C\left( {4,3} \right) + C\left( {4,4} \right)\,\, = \,\,{{4 \cdot 3} \over 2} + 4 + 1\,\, = \,\,11\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle

\left( {\rm{B}} \right)\,\,\,N = 5\,\,\,\,\, \Rightarrow \,\,\,\,B\left( 5 \right)\,\, = \,\,C\left( {5,2} \right) + C\left( {5,3} \right) + C\left( {5,4} \right) + C\left( {5,5} \right)\,\, = \,\,2 \cdot \left( {{{5 \cdot 4 \cdot 3} \over {3 \cdot 2}}} \right) + 5 + 1\,\, = \,\,20 + 5 + 1\,\,\,\,\,\,\,\left\langle {{\rm{YES}}\,{\rm{!}}} \right\rangle

The correct answer is therefore (B).

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

omahawar · Answer

Each bulb can be either ON or OFF.
Total number of posibilities of all Bulbs = 2^N
Total number of ways in which room is NOT well lit = (Only 1 Bulb is ON) + (All Bulbs are OFF) = (N) + 1;

Total number of ways room is well lit = 2^N - (N + 1) = 26

PSKhore · Answer

Total possible combinations:    2^N   because each of the N bulbs has 2 states (on or off).

Number of ways to have the room not well lighted:

Zero bulbs on:   There is only one way to have zero bulbs on (all switches are off).

Exactly one bulb on:   There are N ways to have exactly one bulb on (one for each of the N bulbs).

So, the total number of ways the room is not well lighted is   1+N  .

Number of ways (well lighted) = (Total combinations) - (Ways not well lighted)
  26=2^N − (1+N)

Plug in the options,  5  satisfies
 Hence, B

sashankmvv1234 · Answer

Isn't the bulbs identical? Dont't the solution above think they are different?

For eg. identical bulbs mean A,A,...
For this : AA,AAA,AAAA, etc would be counted differently as compared to seelcting 2 bulbs from n different bulbs..

PSKhore · Answer

Hey, their switches are different. Since each will have a different switch, doesn’t matter if they are identical.

Adriana´s room contains N identical light bulbs with independent switc

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Adriana´s room contains N identical light bulbs with independent switc

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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