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12. Simplify, and find the value of [tex][tex]$\frac{9!}{7!}$[/tex][/tex] without using a calculator.
Answer: [tex]$\qquad$[/tex]

13. In how many ways can six different books be arranged on a shelf?
Answer: [tex]$\qquad$[/tex]

14. How many arrangements of the letters MOTHER can be made taking four letters at a time?
Answer: [tex]$\qquad$[/tex]


Sagot :

Sure, let's solve each of these parts step-by-step.

### Question 12
Simplify, and find the value of [tex]\(\frac{9!}{7!}\)[/tex].

First, let's recall the definition of factorial:
- [tex]\( n! \)[/tex] (n factorial) is the product of all positive integers from 1 to [tex]\( n \)[/tex]. For example, [tex]\( 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)[/tex] and [tex]\( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)[/tex].

The expression [tex]\(\frac{9!}{7!}\)[/tex] can be simplified by canceling out the common factorial terms in the numerator and the denominator:

[tex]\[ \frac{9!}{7!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{9 \times 8 \times (7!)}{7!} = 9 \times 8 = 72 \][/tex]

Answer: [tex]\(\boxed{72}\)[/tex]

### Question 13
In how many ways can six different books be arranged on a shelf?

When considering the arrangement (or permutation) of six different books, we use the factorial of the number of books. The number of ways to arrange 6 different books is [tex]\(6!\)[/tex]:

[tex]\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \][/tex]

Answer: [tex]\(\boxed{720}\)[/tex]

### Question 14
How many arrangements of the letters MOTHER can be made taking four letters at a time?

To find the number of arrangements of four letters taken from the six letters in the word "MOTHER" (where all letters are unique), we need to use combinations and permutations.

First, we select 4 letters out of 6:

[tex]\[ \text{{Number of ways to choose 4 out of 6}} = \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} \][/tex]

Calculating the combinations:

[tex]\[ \binom{6}{4} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (2 \times 1)} = \frac{720}{24 \times 2} = \frac{720}{48} = 15 \][/tex]

Now, each of these 4-letter sets can be arranged in [tex]\(4!\)[/tex] ways:

[tex]\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \][/tex]

Therefore, the total number of arrangements is:

[tex]\[ \binom{6}{4} \times 4! = 15 \times 24 = 360 \][/tex]

Answer: [tex]\(\boxed{360}\)[/tex]
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