Kristian Ferrell
The word basketball contains 10 letters, with some letters repeating. To find the number of distinguishable permutations of the word basketball, we must consider the number of unique arrangements of the letters. Since the letters A, B, and L appear twice, we need to divide the total number of permutations by 2 for each repeated letter. This gives us 10! / 2! 2! 2!, which equals 181,440 distinguishable permutations.
| Characteristics | Values |
|---|---|
| Number of letters in the word BASKETBALL | 10 |
| Number of letters that appear twice | 3 (A, B, and L) |
| Formula for distinguishable permutations | n! / (p! * q! * r!) or 10! / (2! * 2! * 2!) |
| Number of distinguishable permutations | 181,440 |
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What You'll Learn
- The word 'basketball' has 10 letters
- The formula for finding distinguishable permutations is 10! / 2! 2! 2
- The letters B, A, and L appear twice
- The number of arrangements can be calculated by multiplying the number of ways to choose positions for each letter group
- The answer is the product of 605 and 32
The word 'basketball' has 10 letters
The word "basketball" has 10 letters, but when calculating the number of distinguishable permutations, we must also consider that some letters are repeated. Indeed, the word "basketball" has two instances of the letters "A", "B", and "L".
When calculating the number of distinguishable permutations of a word, we typically use a formula that accounts for the total number of letters and the number of each repeated letter. In this case, we have 10 letters in total, with three letters repeated twice. This information is crucial for determining the number of unique arrangements possible.
The formula for calculating distinguishable permutations in this scenario involves considering the various combinations of positions that the repeated letters can occupy. For example, we calculate the number of ways to choose two positions out of eight for the two "A" letters, then the number of ways to choose two positions out of the remaining six for the two "B" letters, and so on. This approach ensures that we account for all the possible arrangements while keeping the repeated letters together.
By applying this strategy, we can determine the total number of distinguishable permutations of the word "basketball." The calculations involve using combinations and factorials to count the different arrangements while ensuring that the repeated letters are not interchangeable without consequence. This process guarantees an accurate count of unique permutations.
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The formula for finding distinguishable permutations is 10! / 2! 2! 2
The formula for finding distinguishable permutations is an important concept in mathematics, and it can be applied to various scenarios, including the word "basketball." Let's delve into the formula and understand how it works.
Now, let's apply this knowledge to the word "basketball." The word "basketball" has 10 letters, but some letters are repeated. There are two As, two Bs, and two Ls. This is where the formula comes into play. By using the formula 10! / 2! 2! 2!, we can find the number of distinguishable permutations.
First, we calculate 10!, which equals 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, resulting in 3,628,800. Next, we calculate 2! 2! 2!, which equals 2 x 1 x 2 x 1 x 2 x 1, resulting in 12. Finally, we divide 3,628,800 by 12, giving us 302,400. This means there are 302,400 distinguishable permutations of the letters in the word "basketball."
The formula 10! / 2! 2! 2! accounts for the repeated letters in the word "basketball." By dividing the total number of permutations by the factorials of the repeated letters, we ensure that we don't overcount arrangements that treat the repeated letters as interchangeable. This is a fundamental principle in permutation calculations.
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The letters B, A, and L appear twice
The word "basketball" has 10 letters, but the number of permutations is less than 10 because there are two of each of the letters B, A, and L. To calculate the number of distinguishable permutations, we can use the formula for distinguishable permutations with repeated elements:
> n! / (n1! * n2! * n3!...), where n is the total number of letters and n1, n2, n3... are the number of each repeated letter.
In this case, n = 10, and there are three letters that appear twice (B, A, and L), so n1 = n2 = n3 = 2. Plugging these values into the formula, we get:
10! / (2! * 2! * 2!) = 3,628,800
So, there are 3,628,800 distinguishable permutations of the word "basketball" where the letters B, A, and L appear twice each.
This problem can also be solved by understanding that the number of distinguishable permutations is calculated by multiplying the number of permutations of the whole set by the number of permutations of each repeated subset. In this case, there are 10! permutations of the whole set of letters in "basketball." For the subsets of B, A, and L, there are 2! permutations of each letter, resulting in (2! * 2! * 2!) permutations of the repeated letters. Multiplying these together gives us the same answer as above:
10! * (2! * 2! * 2!) = 3,628,800
This method accounts for the fact that the repeated letters are indistinguishable from each other, so the total number of permutations is reduced accordingly.
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The number of arrangements can be calculated by multiplying the number of ways to choose positions for each letter group
The word "basketball" has 10 letters, including two Bs, two As, and two Ls. To calculate the number of distinguishable permutations, we can use the formula for permutations when there are repeated letters:
$$\co: 3>\frac{n!}{r_1!\times r_2! \times ... \times r_k}!$$
Where n is the total number of letters, and $r_1, r_2, ..., r_k$ are the number of each repeated letter. In this case, $n = 10$, and there are two letters ($r_1 = 2) of three types (B, A, and L). Plugging these values into the formula, we get:
$$\frac{10!}{2! \times 2! \times 2!} = \frac{3,628,800}{4} = 907,200$$
So, there are 907,200 distinguishable permutations of the letters in the word "basketball."
Now, let's focus on the part about calculating the number of arrangements by multiplying the number of ways to choose positions for each letter group. We can break down the word "basketball" into its letter groups: 2 Bs, 2 As, 2 Ls, 1 K, 1 E, and 1 T. To calculate the number of arrangements, we need to choose positions for each letter group in a way that ensures each arrangement is unique.
For the first B, there are 10 choices (any of the 10 positions in the word). Once the first B is placed, there are only 9 positions left for the second B. So, we multiply: 10 x 9 = 90. This represents the number of ways to choose positions for the two Bs as a group.
Similarly, for the two As, there are 8 choices for the first A (any of the remaining 8 positions after placing the Bs), and then 7 choices for the second A. So, we multiply: 8 x 7 = 56. This gives us the number of ways to choose positions for the two As as a group.
Following the same pattern, we calculate the number of ways to choose positions for the other letter groups:
- Two Ls: 6 x 5 = 30
- One K: 4 (any of the remaining 4 positions)
- One E: 3 (any of the remaining 3 positions)
- One T: 2 (any of the remaining 2 positions)
Now, to find the total number of arrangements, we multiply the number of ways for each letter group:
90 x 56 x 30 x 4 x 3 x 2 = 1,512,000
So, there are 1,512,000 ways to arrange the letter groups in the word "basketball." This method ensures that each arrangement is unique and distinguishable, as it accounts for the positions of the repeated letters.
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The answer is the product of 605 and 32
The word "basketball" contains 10 letters, but since there are two "b's", two "a's", and two "l's", the number of distinguishable permutations is not simply 10 factorial (10!).
Instead, we must account for the repeated letters. One way to do this is to divide by the factorial of the number of each repeated letter. For example, for the two "b's", we divide by 2! because there are two "b's" that can be interchanged without creating a distinguishable permutation. Similarly, we divide by 2! for the two "a's" and 2! for the two "l's". So, the formula for the number of distinguishable permutations is 10! / (2! * 2! * 2!).
This can be simplified further using the formula for the product of 605 and 32. The formula for the number of distinguishable permutations is the product of 605 and 32, which is equal to 19,360. This means there are 19,360 distinguishable permutations of the letters in the word "basketball".
To understand why the formula for the product of 605 and 32 works, let's break it down. The number 605 comes from the fact that there are 6 possible positions for the first "b" (since it can't be in the same position as the second "b"), 5 possible positions for the second "b" (since it can't be in the same position as the first "b"), and 32 possible arrangements of the remaining 8 letters (which include the two "a's" and two "l's"). The number 32 comes from the number of arrangements of the 8 remaining letters, taking into account the two "a's" and two "l's". So, by multiplying 605 by 32, we are essentially counting all the possible arrangements of the letters in "basketball" while avoiding overcounting by accounting for the repeated letters.
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Frequently asked questions
There are 10 letters in the word basketball.
The letter A appears twice in the word basketball.
There are 181,440 distinguishable permutations in the word basketball. This is calculated using the formula n! / (p! * q! * r!), where n is the number of letters, and p, q, and r are the number of times each repeated letter appears.
Yes, the permutations must start with the letter B and end with the letter L.
