Problems about Counting

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Problems about Counting
- Level 1 problems
- Level 2 problems

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These exercises are based on the theory treated on the page Counting

If a problem is solved. It is not 'the' answer.
No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.

Problems about Counting

Level 1 problems

In how many ways can you stack 7 different books, so that a specific book B is on the third place?

Remove book B. Now you can arbitrary stack the 6 books in 6! ways. Afterwards, put book B on the third place.
In how many ways can you take 3 marbles out of a box with 15 different marbles?

Taking a subset of three elements out a set of 15 elements can be done in C(15,3) ways
In a firm are 20 workmen and 10 employees. In how many ways can you construct a delegation with 3 workmen and 2 employees?

First we choose the 3 workmen. This can be done in C(20,3) ways.
Then we choose the 2 employees. This can be done in C(10,2) ways.
The delegation can be assembled in C(20,3).C(10,2) ways
In how many ways can you take 5 cards, with at least 2 aces, out of a game of 52 cards?

First consider 5 cards, with exactly 2 aces.
For the two aces we have C(4,2) ways and for the three other cards we have C(48,3) ways. The 5 cards can be chosen in C(4,2).C(48,3) ways.
Analogous, 5 cards, with exactly 3 aces can be chosen in C(4,3).C(48,2) ways. 5 cards, with exactly 4 aces can be chosen in 48 ways. Total = C(4,2).C(48,3) + C(4,3).C(48,2) + 48
In how many ways can you split a group of 13 persons in 3 persons and 10 persons?

It is sufficient to choose 3 persons to split the group. This can in C(13,3) ways.
How many diagonals are there in a convex n-polygon?

In each vertex there start (n-3) diagonals and since there are n vertices, there start n.(n-3) diagonals. But now we have counted twice each diagonal.
Hence , there are n(n-3)/2 diagonals in a convex n-polygon.
How many numbers consisting of 3 digits, can you make with the digits 0,1,2,3,4 ?

There are 4 ways to choose the first digit of the number.
There are 5 ways to choose the second digit of the number.
There are 5 ways to choose the third digit of the number.
Total = 4.5.5
How many subsets are there in a set S of 10 elements?

We can construct a subset of S by deciding for each element of S, whether or not it belongs to the subset. So, for each element of S, there are 2 possibilities.
Total possibilities = 2¹⁰

Calculate the term with x² in the expansion of (x³ + 1/2x)¹⁰

 
In the expansion of (x³  + 1/2x)¹⁰ we have :

  The first term contains x³⁰
  The second term contains x²⁶
  The third term contains x²²
  ...
  The eighth term contains x²

This eighth term is  C(10,7).(x³)³.(1/(2x))⁷
The coefficient is    C(10,7)/2⁷  = 15/16

1011011101 is an example op a binary number of length 10.
How many binary numbers of length 10 end up with 111 and contain exactly two zeros.

The number is completely defined if we know the places of the 2 zeros.
The number starts with 1 and ends up with 111.
There are still six digits to be determined.
Once we identify two locations for the zeros, the number is fixed.
The number of ways to choose 2 places out of the 6, is C(6,2).

Level 2 problems

In how many ways can you stack m identical bricks into k boxes so that each box contains at least 1 brick.
Since each box must contain at least one brick, we first put one brick in each box. Now, we have to find the number of ways you can stack m-k identical bricks into k boxes.
We describe one way to stack the (m-k) identical bricks into the k boxes.
Number the boxes.
Write on each brick the number of a destination box and order the (m-k) bricks, in ascending order, in one row.
The row of numbers on the bricks constitute a combination with repetition of k different elements (the k numbers) choose m-k (the bricks).
This can be done in
```
 
                                   (m - 1)!
        C'(k,m-k) = C(m-1,m-k) = -----------------     ways.
                                 (m - k)!(k - 1)!
```

From a deck of cards we take 12 cards.

hearts 1, 2 and 3
clubs 1, 2, 3 and 4
diamond 1, 2, 3, 4 and 5

Take 5 cards such that there is at least one card of each type. In how many ways is that possible? The order of these five cards is irrelevant.

How many strictly positive integer solutions ( x, y , z) are there,
such that x + y + z = 100

This is the same problem as a previous problem :
In how many ways can you arrange 100 identical bricks into 3 boxes so that each box contains at least one brick.
With the method of this previous problem, we get the answer C(99,97) = 4851 ways
How many terms are contained in (a + b + c)²⁰ .

All terms can be written as A.a^p .b^q .c^r with p + q + r = 20.
The number of terms is the number of solutions of the equation p + q + r = 20 with p, q, r as positive integer unknowns.
Now regard (p,q,r) as three ordered elements. Point 20 times to one of these elements, and order these elements in the same order as the given elements. This corresponds with one solution of p + q + r = 20 and it is a combination with repetition of 3 elements choose 20.
The number of terms is the number of such combinations
= C'(3,20) = C(22,20) = C(22,2) = 231
The term A.a¹⁰ b³ c⁷ is present in the development of (a + b + c)²⁰
Calculate A.
The number of terms a¹⁰ b³ c⁷ is the number of permutations with repetition of the elements a,a,a,a,a,a,a,a,a,a,b,b,b,c,c,c,c,c,c,c This number is
```
 
                20!
            ----------- = 22 170 720
            10! 3! 7!
```

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