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s344n2d4d5 [400]
2 years ago
9

How many 5-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5, 6, if repetition of digits is not allowed?

Mathematics
1 answer:
shepuryov [24]2 years ago
8 0

answer

c) 2520

Step-by-step explanation:

1st digit 1 of 7

2nd digit 1 of 6

3rd digit 1 of 5

4th digit 1 of 4

5th digit 1 of 3

7*6*5*4*3=2520 combination

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PLZ ANSWER THANK YOU
NikAS [45]

Answer:

1, x/3, x^2/9, x^3/27,  x^4/81


x^4 +3x^3+9x^2+27x+81

----------------------------------------

81


Step-by-step explanation:

(x/3) ^i

i=0   (x/3)^0 = 1

i=1   (x/3)^1 = x/3

i=2   (x/3)^2 = x^2/3^2 = x^2/9

i=3   (x/3)^3 = x^3/3^3 =x^3/27

i=4   (x/3)^4 = x^4/3^4 =x^4/81

The sum is

1+(x/3) + x^2/9 + x^3/27 + x^4/81

We need a common denominator of 81

1*81/81 + x/3 *27/27 + x^2/9 *9/9 + x^3/27 *3/3 + x^4/81

81+27x + 9x^2 + 3x^3 +x^4

----------------------------------------

81

Rewriting from largest power to smallest power

x^4 +3x^3+9x^2+27x+81

----------------------------------------

81


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