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Anika [276]
3 years ago
14

Use the digits 0-9, find out how many 4 digit numbers can be configured based on the stated conditions:

Mathematics
1 answer:
Alexeev081 [22]3 years ago
3 0
A) On the first place we can put 9 digits, on the second 9, on the third place 8 and on the last place 7 ( because no digits can be repeated ):
9 · 9 · 8 · 7 =4536 numbers
b) There are 5 odd digits can be repeated. So:
5 · 10 · 10 · 5 = 2500 numbers
c) On the first place we can put:5,6,7,8,9 (5 digits) On the last place is just one (0), because the number must be divisible by 10:
5 · 10 · 10 · 1 = 500 numbers
d) on the first place we can put 2 digits ( 1, or 2 - there are no 4-digit numbers starting with 0). The last digit must be even, but the order can be:
X - odd - odd - even           2 · 5 · 4 ·  5 = 200
X - odd - even - even         2 · 5  · 5 · 4 = 200
X - even - odd - even         2 · 5  · 5 · 4 = 200
X - even - even - even       2 · 5 ·  4 · 3 = 120 
Finally: 200+ 200 + 200 + 120 = 720 numbers
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\sum_{k=3}^5( - 2k + 5) =  - 9

EXPLANATION

The given series is

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The expanded form is

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This simplifies to

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