I am having trouble with this question as I am not sure I am approaching it in the correct manor.
A so-called 7-way lamp has three 60-watt bulbs which may be turned on one or two
or all three at a time, and a large bulb which may be turned to 100 watts, 200 watts
or 300 watts. How many different light intensities can the lamp be set to give if the
completely off position is not included?
Firstly I am not too sure what the ref to a seven way light bulb is from what I have google I have assumed it means that the intensity of the bulb can be varied with 7 different intensity's and I am assuming that as the bulbs can be turned on separately then the intensity for each can be varied individually.
So my working are as follows
If I think of a single bulb on it own, the I am going to have $^7C_1=7$ different intensities.
Now if I switch on another bulb so now I would have $^7C_1*^7C_1=49$
a brief reasoning for this is as follows:
$B_n=n^{th}$ Bulb
$I_n=n^{th}$ Intensity
If I set $B_1$ to an intensity of $I_1$ then turned on $B_2$ and adjust the intensity of $B_2$ from $I_1$ to $I_7$ then I would form the following combinations
$I_1,I_1$
$I_1,I_2$
.
.
.
.
$I_1,I_7$
So if I apply this to all for 2 bulbs I would get $49$ different intensities.
Applying same logic again to all three $60W$ bulbs then I would have 343 different intensities.
Now if I then add the $100W$ bulb to this I would get $343+7=350$ different intensities.
If I then apply this to varying the bulb power for $200W$ and $300W$ then the number of intensity's I would get now would be:
$$No.intensity=343+7*3=364$$
Iv tried to think of it as I would be operating the switch myself, but am just not 100% that this is correct, is there a flaw with my working or how I have approached this?
Best Answer
Ok, so if there are n devices with k positions, there are $k^n$ possibilities. The possible intensities for the first three bulbs are $0$, $60$, $120$ and $180$ (4 possibilities). The three-way bulb has 3 positions and a off position, so 4 positions in total. So there are $4\cdot 4 = 16$ possibilities. You remove the position completely off and you get $15$ positions.