Mẹo How many permutations are there of the following word? bookkeeping
Thủ Thuật về How many permutations are there of the following word? bookkeeping Chi Tiết
Lê Mạnh Hùng đang tìm kiếm từ khóa How many permutations are there of the following word? bookkeeping được Update vào lúc : 2022-12-12 21:14:06 . Với phương châm chia sẻ Mẹo về trong nội dung bài viết một cách Chi Tiết Mới Nhất. Nếu sau khi Read tài liệu vẫn ko hiểu thì hoàn toàn có thể lại Comments ở cuối bài để Tác giả lý giải và hướng dẫn lại nha.Method 1: We arrange the five letters B, K, K, P, R in a row, then insert the two O's and three E's in that order.
Nội dung chính Show- How many permutation are there in the word?How many distinct permutations can be formed from the word permutation?How many permutation can be made from the words assessments?How many distinct permutations can be formed using the letters of the word baseball?
The number of ways we can arrange the five letters B, K, K, P, R in a row is $$frac5!2!$$ where we divide by $2!$ since we can permute the two K's within a given arrangement without producing an arrangement distinguishable from that arrangement.
We now have six spaces to fill, four between successive letters and two the ends of the row. We can either place both O's in one of these six spaces, which can be done in $binom61$ ways, or place them in two different spaces, which can be done in $binom62$ ways.
Case 1: We place both O's in one of the six spaces.
This creates an arrangement with seven letters in which the two O's are adjacent. Since we are not permitted to have two adjacent O's, we must place an E in between the two O's. We now have an arrangement of eight letters, including the sequence OEO. This creates nine spaces, seven between successive letters and two the ends of the row. Since the E's cannot be adjacent, we must separate them by placing an E in two of the seven spaces that are not adjacent to the E we have placed between the two O's. We can do this in $binom72$ ways. The number of such arrangements is $$binom61binom72$$
Case 2: The two O's are placed in two different spaces.
This creates an arrangement of seven letters in which the O's are not adjacent. We have eight spaces to fill, six between successive letters and two the ends of the row. To separate the E's, we choose three of these eight spaces in which to insert an E, which can be done in $binom83$ ways. The number of such arrangements is $$binom62binom83$$
Total: Since the two cases are disjoint, the number of arrangements of BOOKKEEPER in which the E's are not adjacent and the O's are not adjacent is $$frac5!2!left[binom61binom72 + binom62binom83right] = 57960$$
Method 2: We use the Inclusion-Exclusion Principle.
There are $$frac10!2!3!$$ distinguishable arrangements of the word BOOKKEEPER. From these we must exclude those arrangements in which two O's or two E's are adjacent.
One pair of adjacent identical letters:
Two O's are adjacent: We have nine objects to arrange: B, OO, K, K, E, E, E, P, R. Since the two K's are indistinguishable and the three E's are indistinguishable, they can be arranged in $$frac9!2!3!$$ distinguishable ways.
Two E's are adjacent: We have nine objects to arrange: B, O, O, K, K, EE, E, P, R. Since the two O's are indistinguishable and the two K's are indistinguishable, they can be arranged in $$frac9!2!2!$$ distinguishable ways.
Two pairs of adjacent identical letters:
Two O's are adjacent and two E's are adjacent: We have eight objects to arrange: B, OO, K, K, EE, E, P, R. Since the two K's are indistinguishable, they can be arranged in $$frac8!2!$$ distinguishable ways.
Two pairs of E's are adjacent: We have eight objects to arrange: B, O, O, K, K, EEE, P, R. Since the two K's are indistinguishable and the two O's are indistinguishable, they can be arranged in $$frac8!2!2!$$ distinguishable ways.
Three pairs of adjacent identical letters:
Two O's are adjacent and two pairs of E's are adjacent: We have seven objects to arrange: B, OO, K, K, EEE, P, R. Since the two K's are indistinguishable, they can be arranged in $$frac7!2!$$ distinguishable ways.
By the Inclusion-Exclusion Principle, the number of permissible arrangements is $$frac10!2!2!3! - frac9!2!3! - frac9!2!2! + frac8!2! + frac8!2!2! - frac7!2! = 57960$$
Number of letters in the word BOOKKEEPING = 11.
There are three doubles and 5 singles.
We can take 3 cases to use 5 letters.
Case-1 Taking 2 doubles and 1 other.
Number of permutaions formed by 2doubles and 1 other 
Case-2 Taking 1 double and 3 others.
Number of permutaions formed by 1double and 3 others 
Case-3 Taking 5 others.
Number of permutaions formed by 5 distinct letters 
Total number of perputations formed by using 5 letters of the word BOOKKEEPING is 540 + 6300 + 6720 = 13560.
How many permutation are there in the word?
To calculate the amount of permutations of a word, this is as simple as evaluating n! , where n is the amount of letters. A 6-letter word has 6! =6⋅5⋅4⋅3⋅2⋅1=720 different permutations.How many distinct permutations can be formed from the word permutation?
1 Answer. The Reqd. No. of Permutations=3360.How many permutation can be made from the words assessments?
Answer. Answer: 3 628 800 permutations of letters.How many distinct permutations can be formed using the letters of the word baseball?
Observe that the number of letters in the word BASEBALL is 8. Therefore, n = 8. Therefore, the possible number of permutations is 5040. Tải thêm tài liệu liên quan đến nội dung bài viết How many permutations are there of the following word? bookkeeping
Post a Comment