22**. In how many ways can you make up a ruble using coins of 1, 2, 5, 10, 20, and 50 kopecks?
22. Since from coins of 10, 20, and 50 kopecks, only a whole number of tens of kopecks can be formed, then from 1, 2, and 5 kopeck coins, a whole number of tens of kopecks must also be formed. Therefore, the following cases are possible:
1) The ruble is composed of coins worth 10, 20, and 50 kopecks,
2) Coins worth 10, 20, and 50 kopecks form 90 kopecks.

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-122.jpg?height=730&width=848&top_left_y=454&top_left_x=135)

Let's consider each of these possibilities separately.

1) One ruble can be composed of coins worth 10, 20, and 50 kopecks in the same number of ways as 10 kopecks can be composed of coins worth 1, 2, and 5 kopecks. According to problem 21b), this number is

$$
\left(\frac{(10+4)^{2}}{20}\right)=\left(\frac{196}{20}\right)=(9.8)=10
$$

2) 90 kopecks can be composed of coins worth 10, 20, and 50 kopecks in the same number of ways as 9 kopecks can be composed of coins worth 1, 2, and 5 kopecks, i.e., $\left(\frac{(9+4)^{2}}{20}\right)$ ways. This number is

$$
\left(\frac{169}{20}\right)=(8.45)=8
$$

In this case, we need to form 10 kopecks with coins worth 1, 2, and 5 kopecks. The number of different ways to do this is

$$
\left(\frac{(10+4)^{2}}{20}\right)=\left(\frac{196}{20}\right)=10
$$

Therefore, the total number of different ways to form 90 kopecks with coins worth 10, 20, and 50 kopecks, and 10 kopecks with coins worth 1, 2, and 5 kopecks, is
$8 \cdot 10=80$.

3) By similar reasoning, the total number of different ways is

$$
\begin{aligned}
\left(\frac{(8+4)^{2}}{20}\right) \cdot\left(\frac{(20+4)^{2}}{20}\right) & =\left(\frac{144}{20}\right) \cdot\left(\frac{576}{20}\right)= \\
& =(7.2) \cdot(28.8)=7 \cdot 29=203
\end{aligned}
$$

4) The total number of different ways is

$$
\begin{aligned}
\left(\frac{(7+4)^{2}}{20}\right) \cdot\left(\frac{(30+4)^{2}}{20}\right) & =\left(\frac{121}{20}\right) \cdot\left(\frac{1156}{20}\right)= \\
& =(6.05) \cdot(57.8)=6 \cdot 58=348
\end{aligned}
$$

5) The total number of different ways is

$$
\begin{aligned}
\left(\frac{(6+4)^{2}}{20}\right)\left(\frac{(40+4)^{2}}{20}\right) & =\left(\frac{100}{20}\right) \cdot\left(\frac{1936}{20}\right)= \\
& =(5)(96.8)=5 \cdot 97=485
\end{aligned}
$$

6) The total number of different ways is

$$
\begin{aligned}
\left(\frac{(5+4)^{2}}{20}\right)\left(\frac{(50+4)^{2}}{20}\right) & =\left(\frac{81}{20}\right)\left(\frac{2916}{20}\right)= \\
& =(4.05)(145.8)=4 \cdot 146=584
\end{aligned}
$$

7) The total number of different ways is

$$
\begin{aligned}
\left(\frac{(4+4)^{2}}{20}\right)\left(\frac{(60+4)^{2}}{20}\right) & =\left(\frac{64}{20}\right)\left(\frac{4096}{20}\right)= \\
& =(3.2)(204.8)=3 \cdot 205=615
\end{aligned}
$$

23-24] PROBLEMS IN COMBINATORICS AND PROBABILITY 123

8) The total number of different ways is

$$
\begin{aligned}
\left(\frac{(3+4)^{2}}{20}\right)\left(\frac{(70+4)^{2}}{20}\right) & =\left(\frac{49}{20}\right)\left(\frac{5476}{20}\right)= \\
& =(2.45) \cdot(273.8)=2 \cdot 274=548
\end{aligned}
$$

9) The total number of different ways is

$$
\begin{aligned}
\left(\frac{(2+4)^{2}}{20}\right) \cdot\left(\frac{(80+4)^{2}}{20}\right) & =\left(\frac{36}{20}\right) \cdot\left(\frac{7056}{20}\right)= \\
& =(1.8) \cdot(352.8)=2 \cdot 353=706
\end{aligned}
$$

10) The total number of different ways is

$$
\begin{aligned}
\left(\frac{(1+4)^{2}}{20}\right)\left(\frac{(90+4)^{2}}{20}\right) & =\left(\frac{25}{20}\right)\left(\frac{8836}{20}\right)= \\
& =(1.25)(441.8)=1.442=442
\end{aligned}
$$

11) The total number of all different ways is

$$
\begin{aligned}
\left(\frac{(0+4)^{2}}{20}\right)\left(\frac{(100+4)^{2}}{20}\right) & =\left(\frac{16}{20}\right)\left(\frac{10816}{20}\right)= \\
& =(0.8)(540.8)=1 \cdot 541=541
\end{aligned}
$$

Thus, the final total number of ways to form a ruble with coins worth 1, 2, 5, 10, 20, and 50 kopecks is:

$$
\begin{gathered}
10+80+203+348+485+584+615+ \\
+548+706+442+541=4562
\end{gathered}
$$