Evaluate P(3,3) help please

Mathematics

2 answers:

Lera25 [3.4K]3 years ago

7 0

Answer:

The value of P(3,3) is 6

Step-by-step explanation:

We have to evaluate P(3,3)

As permutation can be calculated by the formula

$P(n,r)=_r^n\txterm{P}=\frac{n!}{(n-r)!}$

$P(3,3)=_3^3\txterm{P}=\frac{3!}{(3-3)!}$

$=\frac{3!}{0!}=\frac{3!}{1}=3\times 2\times 1=6$

Hence, the value of P(3,3) is 6

scZoUnD [109]3 years ago

5 0

$P(n,k)$ where $n=k$ is equal to $n!$

So, $P(3,3)=3!=6$

Generally, $P(n,k)=\dfrac{n!}{(n-k)!}$

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babunello [35]

Answer:

Question 9: Variables: (smallest) s, q, r (largest)

Question 10: 5 whole numbers (7, 8, 9, 10, and 11)

Step-by-step explanation:

For question nine, there are two given statements... s=q-2 and q<r. Say we plug in 10000 (a really big #) in for q, then we would get s=9998 and r>10000. This way, we can see that s would be the smallest, then q, and r is the largest. <em>(q<r can be written as r>q)</em>

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For question 10, it states $\frac{1}{4}$ . This can be split into $\frac{3}{x}$ and $\frac{3}{x}$ . When x is 12 in the first equation then $\frac{3}{12} = \frac{1}{4}$ and when x is 6 in the second equation $\frac{3}{6} =0.5$ (0.5 is also $\frac{1}{2}$ ). Therefore, x must be a whole number less than 12 and greater than 6, and it cannot be either 12 or 6. Whole numbers between 6 and 12 are 7, 8, 9, 10, and 11 or 5 whole numbers.