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3 years ago

How many five digit numbers can be formed from the digits 1,,3,4,5,6,7 if no digit is repeated

Mathematics

2 answers:

zhuklara [117]3 years ago

6 0

There can be up to 36 i believe

Hope this helps

Viktor [21]3 years ago

3 0

To get all possible permutations of these five spaces, we just multiply these numbers (that is, 8∗7∗6∗5∗4=6,720)

There are 6,720 five digit numbers that can be formed from the digits provided given that no digit may be repeated.

Hope that helped!

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Name an event that would be measured in days

telo118 [61]

Well, things like sporting events, or freak storms are measured in days, or weeks. Depends. For weather, it could be hurricanes, tornado's, or even large fires(not quite weather, but could be)

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3 years ago

Read 2 more answers

WILL GIVE BEST RESPONSE

tino4ka555 [31]

We will use demonstration of recurrences1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation 10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, for all positive integers n>=1
3) let's show that the equation is also true for n+1, n>=1
10 + 30 + 60 + ... + 10(n+1) = 5(n+1)(n + 2)
let be N=n+1, N is integer because of n+1, so we have
10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
so the equation is also true for n+1,
finally, 10 + 30 + 60 + ... + 10n = 5n(n + 1) is always true for all positive integers n.

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3 years ago

If the simple interest on $4000 for 3 years is $720, then what is the interest rate?

solong [7]

Answer:

0.06% annually

Step-by-step explanation:

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2 years ago

Solve the systems of equations by the elimination method: -2x + 2y = 2 -4y - 2x = -22

bekas [8.4K]

Answer:

(3,4)

Step by step explanation:

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3 years ago

Help? I can't seem to understand arithmegic sequence.

katrin2010 [14]

A sequence $\{a_n\}$ is arithmetic if the difference between consecutive terms is some fixed number, regardless of which pair of consecutive terms you pick out of the sequence.

For example, the following sequences are arithmetic:

1, 2, 3, 4, 5, 6, ... (difference = 1)

-25, -20, -15, -10, -5, ... (difference = 5)

2. Carla's sequence is not arithmetic, because the differences between consecutive terms are all different:

13 - 11 = 2

17 - 13 = 4

25 - 17 = 8

She can adjust the sequence by changing the last two numbers to 15 and 17, since this makes the difference fixed:

13 - 11 = 2

15 - 13 = 2

17 - 15 = 2

and so on.

3. The sequence

45, 48, 51, 54, ...

is arithmetic with difference 3 between terms. Recursively, we can write the $n$ th term, $a_n$ , in terms of the previous, $(n-1)$ th term, $a_{n-1}$ :

$a_n=a_{n-1}+3$

By this definition, we can just as easily write the $(n-1)$ th term in terms of the $(n-2)$ th term:

$a_{n-1}=a_{n-2}+3$

Then, substituting this into the previous equation, we have

$a_n=(a_{n-2}+3)+3=a_{n-2}+2\cdot3$

We can continue this process to write $a_n$ in terms of $a_1$ :

$a_{n-2}=a_{n-3}+3\implies a_n=a_{n-3}+3\cdot3$

$a_{n-3}=a_{n-4}+3\implies a_n=a_{n-4}+4\cdot3$

and so on. (You might notice that the subscript of the term on the right side, and the number of 3s being added, together sum to $n$ .) The pattern continues down to

$a_n=a_1+(n-1)\cdot3$