有多少个 8 位字符串至少有两个连续的 0 或 1？

排列是给定数量的组件的许多配置，或者通过一个接一个地选择，其中一部分，或者一次全部选择。它们是将线性序列分配给系列成分的过程。重新排序特定序列或系列的元素的过程也称为重新排序。换句话说，排列一个序列需要列出该序列的所有替代排列。例如，序列 1, 2 可以用两种不同的方式表示：1, 2 或 2, 1。

置换公式

当 n 个元素按特定顺序排列 r 个项目时，排列的数量为，

ⁿP_r = n! / (n – r)!

For example, let n = 5 and r = 2. The number of permutations is ⁵P₂ = 5! / (5 – 2)! = 20.

组合

它被描述为从给定序列中选择一个、两个或几个元素的过程，与它们出现的顺序无关。如果您从仅包含两个元素的系列中选择两个组件，则这些元素的顺序无关紧要。

组合配方

当从序列中的 n 个元素中选择 r 项时，组合的数量为

ⁿC_r = n! / r! (n – r)!

For example, let n = 5 and r = 2, then the number of ways to choose two components from a set of five = ⁵C₂ = 5! / 2! (5 – 2)! = 10.

值得一提的是，如果要从n个分量的集合中得到ar个组合，并且这样的元素可以重复，那么

^n+r−1C_r = ^n+r−1C_n−1.

有多少个 8 位字符串至少有两个连续的 0 或 1？

解决方案：

An 8 bit string could represent all numbers between 0 and 2⁸ = 256.

Two consecutive zeroes can start at position 1, 2, 3, 4, 5, 6 or 7.

Starting at position 1: Strings of the form [0 0 x x x x x x]

Remaining 6 places can be arranged in 2⁶ = 64 ways.

Starting at position 2: Strings of the form [ 1 0 0 x x x x x]

Remaining 5 places can be arranged in 2⁵ = 32 ways.

Similarly, the count for positions 3, 4, 5, 6 and 7 comes to be 32 each.

Number of ways = 64 + 32 + 32 + 32 + 32 + 32 + 32 + 32 – 10

= 246

Also number of strings that start with 2 consecutive 1’s = 246

Thus required value = 246 + 246 = 492

类似问题

问题 1. 有多少长度为 5 的二进制字符串在字符串某处恰好有两个 1？

解决方案：

Note that the order of the bit is not important in this case because we are concerned with the number of ones in the said string and not their order. Thus, we need to apply the concept of combinations to find the required value.

Here, n = 5 and r = 2.

C(5, 2) = $\frac{5!}{2!(5-2)!}$

= 10

So there are 10 bit strings of length 5 with exactly two 1’s in them.

问题 2：如果要从 7 名男性和 6 名女性中选出一个由 5 人组成的委员会，找出其中至少有 3 名男性参加的人数。

解决方案：

At least three men on the committee means we can have either exactly three, four or all five men in the committee.

Number of arrangements when there are 3 men and 2 women on the committee = (⁷C₃ x ⁶C₂) = 525

Number of arrangements when there are 4 men and 1 woman on the committee=  (⁷C₄ x ⁶C₁) = 210

Number of arrangements when there are all 5 men on the committee = (⁷C₅) = 21

Total arrangements = 525 + 210 + 21

= 756

问题 3. 找出元音总是一起出现的单词“LEADING”的字母排列数？

解决方案：

If the vowels are to appear together, they would form a separate letter in the word. Hence we are left with 4 + 1 = 5 letters, which can be arranged in 5! = 120 ways.

Furthermore, there are 3! = 6 ways to arrange the vowels together.

Total number of ways of arranging the letters = 120 x 6 = 720.

问题 4. 找出由 8 个辅音和 5 个元音组成的有 4 个辅音和 3 个元音的单词的数量。

解决方案：

Number of ways of selecting 4 consonants out of 8 and 3 vowels out of 5 = ⁸C₄ x ⁵C₃

= $\frac{8 ×7 ×6 ×5 ×4!}{4! × 4!} × \frac{5 ×4 ×3!}{3! × 2!}$

= 70 × 10 = 700

Number of ways of arranging the 7 letters among themselves = 7! = 5040

Number of words that can be formed = 5040 × 700 = 3528000.

问题 5. 如果不允许重复，“GEEKSFORGEEKS”这个词可以组成多少个四个字母的词？

解决方案：

Since there are 7 different letters in the word ‘GEEKSFORGEEKS’

Required number of words = ⁷P₄

= 7! / 3!

= 840

问题 6. 假设存钱罐中有 20 枚硬币，求存钱罐包含五分硬币、一角硬币和 25 美分硬币的组合数。

解决方案：

Clearly order of the coins does not matter in this question. Also there is no mention as to whether repetition is allowed or not.

Following the rule, there are in case of r number of combinations from a sequence having n number of elements where elements can be repeated ^n+r−1C_r = ^n+r−1C_n−1, we have:

Number of combinations = ^20+3-1C₂₀ = ²²C₂₀

= 22! / (22 – 20)! 20!

= 11 × 21

= 231

问题 7. 假设我们有 1 个三人间和 2 个双人间，请说出分配 7 名学生参加大学旅行的方式的数量。

解决方案：

This problem can be interpreted as having to put the 7 students into groups of 3, 2 and 2.

Number of ways to choose 3 students in the triple = ⁷C₃ = 7! / 3!4! = 35  

Number of ways to choose 2 out of the remaining 4 students = ⁴C₂ = 4!/ 2!2! = 6

Number of ways to choose 2 students out of the remaining two students = 1

Total number of arrangements = 35 × 6 × 1 = 210.

Hence, 7 students can be assigned to 1 triple and 2 double hotel rooms during a conference in 210 ways.