Question 1091481: How many even three-digit numbers can be formed from the
digits: 0, 1, 2, 3, 4, 5, 6 such that 5 is never placed
in the tens position?
Found 2 solutions by Edwin McCravy, Gentle Phill:
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website!
I will assume that the digits can be repeated.
Choose the first (hundreds) digit any of 6 ways. {1,2,3,4,5,6}
Choose the second (tens) digit any of 6 ways. {0,1,2,3,4,6}
Choose the third (units or ones) digit any of 4 ways (0,2,4,6}
answer = 6 = 144 ways.
If the digits cannot be repeated, the problem is more complicated
and requires different cases. The answer would be 88. If you need
that instead, tell me in the thank-you note form below and I'll get
back to you by email, to show how that is done.
Edwin
Answer by Gentle Phill(18)  (Show Source):
You can put this solution on YOUR website! #Soln
.
Given d digits:
0,1,2,3,4,5,6
.
To form even 3-digits without repitition & place of 50:
.
1. Nos must end with either 0,2,4,6
.
2. Nos must not Bgin with 0
.
3. By given rule, 5 must not appear in the middle
.
To ensure coverage for all of the above without contradiction, we must observe D many cases possible bit by bit
.
#Case_1: 0 as unit
.
When 0 is unit, any no. can occupy 1st spot:
1_0
2_0
3_0
4_0
5_0
6_0
.
However:
.
~ When 1,2,3,4 nd 6 re 1st digit, 5 is neva used. We'll've:
.
[5][4][1] = 20
.
Since 0 takes last spot, 1 of 5 nos takes 1st spot and 1 of d remainin 4 nos take 2nd spot
.
~ When 5 occupies 1st spot, we'll've:
.
[1][5][1] = 5
.
#Case_2: Either of 2,4,6 as unit
.
Digit could start with 1,3,5 & any other 2 even not used as unit
.
However:
.
~ When 5 isn't 1st digit, it isn't used at all:
.
[4][4][3] = 48
.
i.e
[1,3 or any 2 even not used up][remaining 4 nos excluding '5'][2,4 or 6]
.
~ When 5 is used:
.
[1][5][3] = 15
.
i.e
[only 5][any other 5 excluding used up evens][2,4,6]
.
Total = 20+5+48+15 = 88
.
.
.
Your friend,
Francis.
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