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Let us count up!

Example 1

Three horses, Tornado (T), Scooter (S) and Victory (V) are competing. How many different outcomes can there be, if a tie is also possible?

Solution

Let us systematise the options based on the ties.

Tie between three horses for first place: \latex{1} option.
Tie between two horses can be
- for \latex{1}st place, in this case T or S or V is the second: \latex{3} options;
- or for \latex{2}nd place, in this case T or S or V is the first: \latex{3} options.
No tie. Thus we get the possible orders of finish when following the arrows in Figure 2: 3 × 2 × 1 = 6 options.

\latex{3}rd place\latex{1}st place\latex{2}nd placeVTSSVVSTTVSVTST\latex{3}×\latex{2}×\latex{1}\latex{3}×\latex{2}\latex{3}Figure 2

The horses can finish in a total of \latex{1} + \latex{3} + \latex{3} + \latex{6} = \latex{13} ways.

Example 2

How many positive integers less than \latex{100} are there for which the sum of the digits is \latex{10}?
How many positive integers less than \latex{1,000} are there for which the sum of the digits is \latex{10}?

Solution (a)

Let us systematise the numbers based on the number of digits.

The sum of the digits of a one-digit number cannot be \latex{10}.
The sum of the digits of a two-digit number is 10 = 1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 5 + 5. Using two distinct digits two different two-digit numbers can be created, thus there are 2 × 4 + 1 = 9 suitable two-digit numbers.

Consequently there are \latex{9} suitable numbers all together.

Solution (b)

Let us systematise the numbers based on the number of digits:

The sum of the digits of a one-digit number cannot be \latex{10}.
According to the previous solution there are \latex{9} two-digit numbers.

Three-digit number:
- From the two-digit numbers, where the sum of the digits is \latex{10}, we can easily create suitable three-digit numbers with the help of digit \latex{0}. As we can place the digit \latex{0} to the local value of ones and tens, we can create 2 × 9 = 18 three-digit numbers.
- Henceforth we concentrate on those three-digit numbers which do not contain the digit \latex{0}.

As the sum of three different digits 10 = 1 + 2 + 7 = 1 + 3 + 6 = 1 + 4 + 5 = 2 + + 3 + 5, which give \latex{4} options. Now let us count up how many three-digit numbers there are the digits of which are distinct given numbers (e.g. \latex{1}, \latex{4}, \latex{5}). Digit \latex{1} can be placed to the local value of hundreds, tens and ones, which give \latex{3} options, then digit \latex{4} can only be placed in two places. As for any placement of digit \latex{1} there are two possible placements of digit \latex{4}, these two digits can be placed in 3 × 2 ways in total. Finally there is one place left for digit \latex{5}, thus using three different digits 3 × 2 × 1 = 6 different three-digit numbers can be created. Thus there are 6 × 4 = 24 three-digit numbers which consist of distinct digits, do not contain digit \latex{0}, and the sum of their digits is \latex{10}.

If a digit can be used repetitively, then 10 = 1 + 1 + 8 = 2 + 2 + 6 = 3 + 3 + 4 = = 4 + 4 + 2, which give \latex{4} options. Let us count up how many three-digit numbers there are which have two equal digits but their third digit is a different one (e.g. \latex{1}, \latex{1}, \latex{8}). The three-digit numbers are defined by the place of digit \latex{8}, which can be placed in three places, thus we get \latex{3} different three-digit numbers. Thus there are 3 × 4 = 12 three-digit numbers which contain a digit twice, and the sum of their digits is \latex{10}.
\latex{10} cannot be expressed as the sum of three equal digits.

Summarising the number of options: there are 9 + 18 + 24 + 12 = 63 suitable numbers.

Example 3

We colour the faces of a cube with \latex{3}-unit long edges to red, then with cuts parallel to the faces we cut the cube into cubes with \latex{1}-unit long edges.

What is the minimum number of necessary cuts if we are allowed to cut only one piece into half at once?
What is the minimum number of necessary cuts if we are allowed to cut several pieces at once?
How many unit cubes did we get which do not have red faces, and how many did we get with \latex{1, 2, 3, 4, 5, 6} red faces?

Solution (a)

Let us try! A possible series of cuts can be the following (Figure 3):

a) Firstly with two cuts we slice the cube into \latex{3} pieces of 3 × 3 × 1 prisms.
b) With two cuts we can cut a 3 × 3 × 1 prism into \latex{3} pieces of 1 × 1 × 3 prisms, thus from the \latex{3} pieces of 3 × 3 × 1 prisms with 3 × 2 = 6 cuts we can get 3 × 3 = 9 pieces of<br>1 × 1 × 3 prisms.
c) With two cuts we can cut a 1 × 1 × 3 prism into \latex{3} pieces of unit cube, thus from the \latex{9} pieces of 1 × 1 × 3 prisms with 9 × 2 = 18 cuts we can get \latex{27} pieces of unit cubes.

Therefore 2 + 6 + 18 = 26 cuts are enough.
It is not possible to cut the cube into unit cubes in fewer steps than above as we should see that if we are allowed to cut only one piece into half at once, then the number of pieces increases exactly by one at every cut. As after the cuts of 1 cube we should have \latex{27} unit cubes, we do need \latex{26} cuts.

Solution (b)

Let us try! According to figure \latex{4} we can cut off the cube with \latex{6} cuts corresponding to the conditions. Therefore \latex{6} cuts are sufficient; we are going to show that \latex{6} cuts are also necessary.

During the cutting we get 3 × 3 × 3 = 27 cubes out of the \latex{1} cube. With one cut we are allowed to half all pieces if we want to, thus with one cut the number of pieces increases to double at most. If we managed to do it in every step (except for the last step), then the number of pieces would be \latex{1, 2, 4, 8, 16, 27} respectively, so at least \latex{5} cuts would be necessary. This number of steps is not enough though, because \latex{6} cuts are necessary to free the small cube in the centre as a cut is necessary along all its \latex{6} faces. Therefore if we are allowed to cut off several pieces at once, then \latex{6} cuts are sufficient and also necessary to dissect the cube.

Solution (c)

Let us observe where those unit cubes can be found on the original cube which have the same number of red faces.

\latex{4} or more red faces: none of the unit cubes.
\latex{3} red faces: the unit cubes at the vertices of the original cube so their number is equal to the number of vertices of the cube, thus there are \latex{8} of them.
\latex{2} red faces: the unit cubes in the middle along each edge of the original cube so their number is equal to the number of edges of the cube, thus there are \latex{12} of them.
\latex{1} red face: the unit cubes in the centre of each face of the original cube so their number is equal to the number of faces of the cube, thus there are \latex{6} of them.
\latex{0} red faces: \latex{1} unit cube in the centre of the original cube.

Example 4

We roll a yellow and a red regular die (the faces are numbered from \latex{1} up to \latex{6} and the sum of the numbers on the opposite faces is \latex{7}).

How many numbers can we get if we add the rolled numbers?
In how many cases will the sum be \latex{1, 2, 4, 7, 11, 12} and \latex{13}?
In how many cases will the product of the rolled numbers be even?

Solution

Let us represent the possible results of rolling two distinct dice with

the help of a square which is divided into six equal parts both vertically
and horizontally.

Let the columns respectively mean that we roll \latex{1, 2, 3, 4, 5, 6} with

the yellow die, and the rows below each other that we roll \latex{1, 2, 3, 4, 5, 6}
with the red die.

Thus with the two dice we can roll a total of \latex{36} number pairs, and
for example the fourth square in the second row means that we rolled
\latex{4} with the yellow and \latex{2} with the red die.

Into the small squares in figure 6 we put the sum, and in figure \latex{7} we put
the product of the rolled numbers. We can read the answers from
the Figures.

\latex{2}\latex{3}\latex{4}\latex{5}\latex{6}\latex{7}\latex{3}\latex{4}\latex{5}\latex{6}\latex{7}\latex{8}\latex{4}\latex{5}\latex{6}\latex{7}\latex{8}\latex{9}\latex{5}\latex{6}\latex{7}\latex{8}\latex{9}\latex{10}\latex{6}\latex{7}\latex{8}\latex{9}\latex{10}\latex{11}\latex{7}\latex{8}\latex{9}\latex{10}\latex{11}\latex{12}

Figure 6

The sum of the rolled numbers can be an integer not less than \latex{2} and not greater than \latex{12}; there are \latex{11n integers like this (Figure 6).

From Figure 6 we can read the number of the correct small squares: the sum is \latex{1}: not possible; the sum is \latex{2}: \latex{1} way; the sum is \latex{4}: \latex{3} ways; the sum is \latex{7}: \latex{6} ways; the sum is \latex{11}: \latex{2} ways; the sum is \latex{12}: \latex{1} way; the sum is \latex{13}: not possible.

Solution I: In Figure 7 we coloured those squares green for which the product of the rolled numbers is even; there are \latex{27} small squares like this, thus the product of the rolled numbers can be even in \latex{27} ways.

Solution II: The product is even if it is not odd. The product is odd
if both of its factors are odd, it is possible in 3 × 3 = 9 ways; as there
are \latex{36} options in total, the product can be even in 36 – 9 = 27 ways.

Solution III: The product of the rolled numbers is even, if we roll an even number with the yellow die or we roll an even number with the red die. It also includes the cases when we roll an even number with both dice. We roll an even number with the yellow die and we roll any number with the red die, it gives 3 × 6 = 18 options. We roll an even number with the red die and we roll any number with the yellow die, it gives 3 × 6 = 18. This would give \latex{36} options in total but we counted those options twice when we roll an even number with both dice, so these 3 × 3 = 9 options should be deducted, thus the product of the rolled numbers can be even in 36 – 9 = 27 ways.

\latex{1}\latex{2}\latex{3}\latex{4}\latex{5}\latex{6}\latex{2}\latex{4}\latex{6}\latex{8}\latex{10}\latex{12}\latex{3}\latex{6}\latex{9}\latex{12}\latex{15}\latex{18}\latex{4}\latex{8}\latex{12}\latex{16}\latex{20}\latex{24}\latex{5}\latex{10}\latex{15}\latex{20}\latex{25}\latex{30}\latex{6}\latex{12}\latex{18}\latex{24}\latex{30}\latex{36}

Figure 7

Exercises

{{exercise_number}}. The five Olympic rings are coloured in five colours: yellow, red, blue, green and black. If we are allowed to colour the rings arbitrarily then how many colourings are possible?

{{exercise_number}}. The following words fell into pieces and the letters are mixed. In how many orders can we write them down? Let us find as many meaningful words as possible.

PAN

PEAR

PANEL

PASTE

PLATES

{{exercise_number}}. Four friends ordered four different cakes in the confectionery: Ann ordered chocolate cake, Barbara ordered chestnut cube, Cissy ordered Japanese cake and Dot ordered fruit cake. The waiter immediately forgot who ordered what, and without asking anything he simply serves the cakes.

In how many different ways can he do it?
In how many ways is it possible that Ann, Barbara and Cissy get what they ordered, but Dot does not?
In how many ways is it possible that only Ann gets what she ordered?
In how many ways is it possible that only one of them gets what she ordered?

{{exercise_number}}. If we write down the natural numbers from \latex{0} up to \latex{2,000} then how many digits do we put down?

{{exercise_number}}. How many pieces of digit \latex{1} did we use when we wrote down the integers from \latex{0} up to \latex{999}?

{{exercise_number}}. Bob has plenty of digits \latex{0, 1, 2, 4, 5, 6, 7, 8, 9}, but he has only a limited number of digit \latex{3}. What is the largest page number he can put down in his notebook using these digits if he has the following number of digit \latex{3}:

\latex{23};

\latex{82};

\latex{181}?

{{exercise_number}}. We put down digits \latex{1, 2, 3, 4} on separate pieces of paper and we put these pieces of paper into a hat. We draw one piece of paper, we jot down the number seen on it, and then we place the piece of paper back. We repeat this action four times in a row thus we get a four-digit number in the end.

How many different four-digit numbers can we get?
How many four-digit numbers can we get in which the first digit is greater than the last digit?
How many four-digit numbers can we get which are divisible by \latex{4}?
How many four-digit numbers can we get which are not greater than \latex{3,000}?

{{exercise_number}}. We write down the digits of \latex{6,174} in all possible orders and then we put these four-digit numbers into ascending order. Which four-digit number will be in the \latex{18th} place?

{{exercise_number}}. We colour a cube with \latex{5}-unit long edges red, then with cuts parallel to the faces we cut the cube into cubes with \latex{1}-unit long edges.

What is the minimum number of necessary cuts if we are allowed to cut several pieces at once?
What is the minimum number of necessary cuts if we are allowed to cut only one piece into half at once?
How many unit cubes do we get which do not have red faces, and how many do we get with \latex{1, 2, 3, 4, 5, 6} red faces?

{{exercise_number}}. How many different bodies can be glued using a) \latex{4}; b) \latex{5} unit cubes, if we exactly align one face of each of two unit cubes at every gluing step?

{{exercise_number}}. How many unit cubes can be created which have

\latex{1} blue face and \latex{5} yellow faces;
\latex{2} blue faces and \latex{4} yellow faces;
\latex{3} blue faces and \latex{3} yellow faces;
\latex{4} blue faces and \latex{2} yellow faces?

Two cubes are not considered to be different if one can be rolled so that we get the other one. E.g. in the figure each of the two cubes has \latex{2} blue faces, and the first cube can be tilted back so that we get a cube corresponding to the second one, thus these two cubes are not considered to be different.

{{exercise_number}}. Alan and Zack are playing according to the following rules: they both roll a die, Alan wins if the two rolled numbers are equal, Zack wins if the two rolled numbers are relative primes. (Two positive integers are relative primes if their greatest common divisor is \latex{1}.) In all other cases it is a tie. From all the possible cases in how many ways can Alan win and in how many ways can Zack win?

{{exercise_number}}. Gladys and Sue are playing according to the following rules: they both roll a die and they add up the rolled numbers. Gladys wins if the sum is a prime number, in all other cases Sue wins. From all the possible cases in how many ways can Gladys win and in how many ways can Sue win?

{{exercise_number}}. Kate and Dan are playing according to the following rules: they both roll a die and they add up the rolled numbers. Kate wins if the sum is at least \latex{6} and at most \latex{8}, in all other cases Dan wins. From all the possible cases in how many ways can Kate win and in how many ways can Dan win?

{{exercise_number}}. Sophie and Dolly are playing according to the following rules: they both roll a die and they multiply the rolled numbers together. Sophie wins if the product is divisible by \latex{5} or is odd, in all other cases Dolly wins. From all the possible cases in how many ways can Sophie win and in how many ways can Dolly win?

{{exercise_number}}. Terry and Steve are playing according to the following rules: they both roll a die and they multiply the rolled numbers together. Terry wins if the product is at most \latex{6} or at least \latex{24}, in all other cases Steve wins. From all the possible cases in how many ways can Terry win and in how many ways can Steve win?

{{exercise_number}}. Let us come up with two-player dice games which are fair, i.e. both players can win in the same number of ways.

{{exercise_number}}. Recently it was raining on \latex{11} days. If it was raining in the morning, then it was sunny in the afternoon; but if it was raining in the afternoon, then it was sunny in the morning. It was sunny for \latex{9} mornings and \latex{12} afternoons in total. On how many days mentioned in the story was there no rain at all?

Puzzle

How many squares can you see in the figure?

Mathematics 9.

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