Can someone help me with these math problems
2 answers:
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Answer:
Step-by-step explanation:
<em>Step 1: Re-state the problem in an easier way to set up a permutation problem</em>
"The numbers 1,6,8,13,15,20 can be placed in the circle below, each exactly once, so that the sum of each pair of numbers adjacent in the circle is a multiple of seven"
The series above can be represented again, in form of:
7a + 1, 7b - 1, 7c + 1, 7d - 1, 7e + 1, 7f - 1
with a, b, c, d, e, f are non-negative integers.
=> 3 numbers are multiply of 7 plus 1
=> 3 numbers are multiple of 7 minus 1
<em>Step 2: Perform the counting:</em>
For the 1st number, there are 6 ways to select
To satisfy that each pair of numbers creates a multiple of 7, then:
For the 2nd number, there are 3 ways left to select
For the 3rd number, there are 2 ways left to select
For the 4rd number, there are 2 ways left to select
For the 5th number, there are 1 way left to select
For the 6th number, there are 1 way left to select
=> In total, the number of possible ways to select:
N = 6 x 3 x 2 x 2 x 1 x 1 = 72
However, these numbers are located around a circle, each option is counted twice.
=> The final number of possible ways:
N = 72/2 = 36
Hope this helps!
:)