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1. Find the number of five-digit multiples of 5, where all the digits are different, and the second digit (from the left) is odd. - This Question was already asked but never answered

 

2. The numbers 1, 2, 3, 4, 5, and 6 are to be entered into the boxes below, so that there is one number per box, and each number is used exactly once. In addition, in each row, the numbers increase from left to right, and in each column, the numbers increase from top to bottom. Find the number of possible placements.

 

 May 4, 2019
 #1
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2- See the solution here: https://web2.0calc.com/questions/homework-problem-and-need-help#r1

 May 4, 2019
 #2
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 May 4, 2019
edited by Guest  May 4, 2019
 #3
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1 - ABCDE:

 

1- For A you have 9 choices.

2 - For B, you have 5 choices if A is even and 4 choices if A is odd.

3- For C, you have 7 choices exlcluding A and B and either a 0 or a 5 which must be at the beginning of the number.

4- For D, you have 6 choices excluding A, B, C and either 0 or 5

5-For E, you ONLY have 2 choices, since the number is a multiple of 5, it must begin with either 0 or 5.

Conclusion: When A is EVEN, you have: 9 x 5 x 7 x 6 x 2 =3,780 - such numbers.

                    When A is ODD, you have  : 9 x 4 x 7 x 6 x 2 =3,024 - such numbers.

                    We add the 2 numbers together =3,780 + 3024 =6,804 - The Total.

 May 4, 2019
 #4
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This answer is incorrect

Guest May 6, 2019
 #5
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I won't look at this question because you have not done these things.

You have instead been thoughtless, lazy and rude.

 May 6, 2019
edited by Melody  May 6, 2019

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