This seems like a extra complicated model of the basic “animal legs” drawback. Except in that case, you solely have two animals to cope with, so you would work it out utilizing elementary maths (“each time I swap a cow for a rooster, I get two extra legs…”).

But on this case you’ve extra variables, so you will have to resort to the algorithmic approach. Recall that within the animal legs drawback, you get one thing like this:

number_of_chickens2 + number_of_cows4 = number_of_legs

number_of_chickens + number_of_cows = number_of_heads

number_of_legs and number_of_heads are recognized, so substitute the equations into one another.

In your instance, there are a lot of extra variables – extra “animals” – which appears to be like one thing like this:

number_of_foo + number_of_nachos + number_of_pizzas2 + number_of_bars3 + number_of_cakes*4 = 24

number_of_foo + number_of_nachos + number_of_pizzas + number_of_bars + number_of_cakes = 8

A typical strategy to clear up that is to precise this as a matrix and carry out row discount on it, the only methodology of which (each by hand and by laptop algorithm) is Gaussian elimination. You might have been taught this in highschool.

| 1 1 2 3 4 | 24 |
| 1 1 1 1 1 | 8  |

(swap row1 and row2)
| 1 1 1 1 1 | 8  |
| 1 1 2 3 4 | 24 |

(row2 = row2 - row1)
| 1 1 1 1 1 | 8  |
| 0 0 1 2 3 | 16 |

(row1 = row1 - row2)
| 1 1 0 -1 -2 | -8 |
| 0 0 1  2  3 | 16 |

Now that the equations are in decreased row echelon kind, you can begin plugging in values to see which of them work. Note that your matrix is not utterly decreased, which tells you there are a lot of doable options.

Let’s begin with row 2: it’s principally saying: number_of_pizzas + number_of_bars*2 + number_of_cakes*3 = 16. Obviously remember that you possibly can’t have damaging numbers of something, which suggests that you would be able to’t have greater than 8 of something. With that in thoughts, we are able to plug in 8 for the variety of bars, which implies 0 pizzas and 0 truffles:

01 + 82 + 0*3 = 16

Now put these values into row 1:

number_of_foo1 + number_of_nachos1 + 00 + 8-1 + 0*-2 = -8

number_of_foo + number_of_nachos = 0

Now we have now our first answer: 8 bars.

Going again to the matrix, since any set of values that satisfies these row equations work, we are able to additionally attempt 5 truffles, which forces us to decide on 1 pizza:

11 + 02 + 5*3 = 16

Into row 1:

number_of_foo1 + number_of_nachos1 + 10 + 0-1 + 5*-2 = -8

number_of_foo + number_of_nachos = 2

Now we have now a second set of options, so long as the variety of foo and nachos complete 2 (they price the identical so it would not matter), and there is 1 pizza and 5 truffles.

Let’s attempt yet one more, simply to get the dangle of it. We can have 4 truffles, and a pair of bars, which leaves 0 pizzas:

01 + 22 + 4*3 = 16

Into row 1:

number_of_foo1 + number_of_nachos1 + 00 + 2-1 + 4*-2 = -8

number_of_foo + number_of_nachos = 2

Again, we have now 2 foo-or-nachos, however this time additionally 2 bars and 4 truffles.

You could also be questioning why we’re guessing the values on the finish – is not there an algorithmic strategy to clear up this? Sure, there’s, it is known as integer programming which is sadly NP-hard. So guessing, or going by means of all the probabilities, must be adequate.