B部分:概率与统计 [41分] 一个编码由10位数字组成,这些数字只能是0或1,例如:1011011010。

  1. (i)

    在无任何限制的情况下,计算此类编码的数量。

  2. (ii)

    若进一步已知这 10 位数字由 4 个 0 和 6 个 1 组成,在没有其他限制的情况下,计算此类编码的数量。

  3. (iii)

    又已知这 10 个数字由 4 个 0 和 6 个 1 组成,如果所有的 0 必须互不相邻,且首位和末位数字必须不同,求这种编码的数量,

  4. (iv)

    又已知这 10 个数字由 4 个 0 和 6 个 1 组成,如果连续的 1 不超过 4 个,求这种编码的数量。

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完整解法

8(i)

1. Apply the multiplication principle to find the total number of unrestricted codes

There are 1010 positions for the digits, and each position has 22 independent choices (either 00 or 11).
Total number of codes=210=1024\text{Total number of codes} = 2^{10} = 1024

答案

1024

8(ii)

1. Calculate the number of arrangements of 10 digits containing 4 identical zeros and 6 identical ones

Number of codes=10!4! 6!(or 10C4)\text{Number of codes} = \frac{10!}{4!\,6!} \quad \text{(or } ^{10}C_4 \text{)}
=210= 210

答案

210

8(iii)

1. Set up the 'slotting' method to ensure all zeros are separated.

To ensure the 44 zeros are separated, we first arrange the 66 ones in a row. Since the ones are identical, there is only 11 way to do this.

_  1  _  1  _  1  _  1  _  1  _  1  _\_ \; 1 \; \_ \; 1 \; \_ \; 1 \; \_ \; 1 \; \_ \; 1 \; \_ \; 1 \; \_

This creates 77 spaces in total (55 internal spaces between the ones, and 22 spaces at the extreme ends) where the zeros can be placed.

2. Translate the condition 'first and last digits must be different'.

Since the digits are only zeros and ones, for the first and last digits to be different, the code must either start with 00 and end with 11, or start with 11 and end with 00.
In terms of the spaces created, this means exactly one of the 22 end spaces must be chosen for a zero.

Number of ways to choose one end space = 2C1=2^2C_1 = 2

3. Place the remaining zeros and calculate the total combinations.

Once one end space is chosen, the remaining 33 zeros must be placed in the 55 internal spaces.

Number of ways to choose the internal spaces = 5C3=10^5C_3 = 10

Total number of such codes = 2×10=202 \times 10 = 20

答案

2020

8(iv)

1. Identify the total number of unrestricted codes (from part ii).

\text{Total unrestricted arrangements} = \frac{10!}{4!6!} = 210

2. Calculate the number of codes where all 6 ones are together.

\text{Treat the } 6 \text{ ones as a single block. We are arranging this } 1 \text{ block and the } 4 \text{ identical zeros.}
\text{Number of arrangements} = \frac{(1 + 4)!}{4!} = \frac{5!}{4!} = 5

3. Calculate the number of codes where exactly 5 ones are together.

\text{Group } 5 \text{ ones into a block, leaving one separate } '1'.
\text{To ensure they do not combine to form } 6 \text{ ones together, they must be separated by at least one } '0'.
\text{The } 4 \text{ zeros create } 5 \text{ spaces between and around them: } _ 0 _ 0 _ 0 _ 0 _
\text{Place the block of } 5 \text{ ones and the separate } '1' \text{ into } 2 \text{ different spaces.}
\text{Number of arrangements} = 5 \times 4 = 20

4. Subtract the unwanted cases from the total to find the final number of valid codes.

\text{Number of codes with no more than } 4 \text{ ones together} = 210 - 5 - 20 = 185

答案

185

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