Evaluate the factorial expression
8!/6!
Use the formula for nPr: How many arrangements can be made using the letters of the word HYPERBOLAS if no letter is to be used more than once.
The numbers 6 and below will cancel each other out leaving 8*7 = 56
To evaluate the factorial expression 8!/6!, we can simplify it using the formula:
n! / r! = (n * (n-1) * ... * (n-r+1))
In this case, we have:
8! / 6! = (8 * 7 * 6!) / 6!
The 6! terms in the numerator and denominator cancel out, leaving us with:
8 * 7 = 56
Therefore, the value of the expression 8!/6! is 56.
Now let's move on to the next question.
To calculate the number of arrangements using the letters of the word HYPERBOLAS without repetition, we can use the formula for the permutation of objects, where n represents the total number of objects and r represents the number of objects to be arranged:
P(n, r) = n! / (n-r)!
In this case, we have a total of 10 letters in the word HYPERBOLAS, and we want to arrange them without repetition. So we have:
n = 10 (total number of letters)
r = 10 (arrange all 10 letters)
Using the formula:
P(10, 10) = 10! / (10-10)!
= 10! / 0!
= 10!
Therefore, there are 10! (10 factorial) arrangements that can be made using the letters of the word HYPERBOLAS without repetition.
To evaluate the factorial expression 8!/6!, we can simplify it by canceling out the common terms in the numerator and denominator.
First, let's expand the expression:
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
6! = 6 * 5 * 4 * 3 * 2 * 1
By canceling out the common terms, we are left with:
8!/6! = (8 * 7) * (6 * 5 * 4 * 3 * 2 * 1)/(6 * 5 * 4 * 3 * 2 * 1)
This simplifies further to:
8!/6! = (8 * 7)/(1) = 56
So, 8!/6! is equal to 56.
Now let's move on to the second part of your question.
To find the number of arrangements that can be made using the letters of the word "HYPERBOLAS" without repeating any letter, we can use the formula for permutations (nPr).
The formula for permutations of "n" objects taken "r" at a time is given by:
nPr = n! / (n - r)!
Applying this formula to our problem, we have:
n = 10 (the number of letters in "HYPERBOLAS")
r = 10 (we want to use all the letters without repetition)
Therefore, the number of arrangements can be given as:
nPr = 10!/(10-10)!
= 10!/0!
= 10!/1
= 10!
So, the number of arrangements that can be made using the letters of the word "HYPERBOLAS" without repetition is 10!.