Question 629274: Different positive four-digit integers are to be formed by using each of the digits 1,2,3,4 just once in each integer. How many different such integers can be formed if the digits 3 and 4 must NEVER be next to each other?
A. 4
B. 8
C. 12
D. 16
E. 24
Found 3 solutions by Edwin McCravy, Theo, ikleyn:
Answer by Edwin McCravy(20081)   (Show Source):
You can put this solution on YOUR website!
The answer would be 4! or 24 ways if it didn't matter whether 3 and 4
were side by side or not. From the 4! we must subtract the ways in which
they are side by side. There are 2 cases where they can be side by side:
Case 1. When the 3 is left of the 4, which means we are arranging these three
things: 1, 2, 34 which is 3! or 6 ways.
Case 2. When the 4 is left of the 3, which means we are arranging these three
things: 1, 2, 43 which is also 3! or 6 ways.
So the answer is 4! - 2x3! = 24 - 2x6 = 24 - 12 = 12 ways.
But, heck, by the time you did all that you could just have listed them all
and counted them:
1. 1324
2. 1423
3. 2314
4. 2413
5. 3124
6. 3142
7. 3214
8. 3241
9. 4123
10. 4132
11. 4213
12. 4231
Edwin
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! i believe the answer is going to be 12.
here's how.
assume that 3 and 4 HAVE to be together.
They are treated as 1, and the formula becomes 3! rather than 4!.
3! equals 6 ways.
we'll let the number 7 represent 3 and 4 together.
the 6 ways are:
127
172
217
271
712
721
now we take the 7 and break it up into A and B so that they can be distinguished easier from the numbers 1 and 2.
we get:
12AB
12BA
21AB
21BA
1AB2
1BA2
2AB1
2BA1
AB12
AB21
BA12
BA21
since 4 number can be arranged 4! ways with no restrictions, then the number of ways we can get the 4 numbers arranged when we can't have 3 and 4 together would have to be 24 - 12 = 12.
those ways would be:
1A2B
1B2A
2A1B
2B1A
A1B2
A2B1
B1A2
B2A1
A12B
B12A
A21B
A12B
just replace the A and B with 3 and 4 and you have your answer in the format the problem was presented.
this works with 3 numbers as well.
suppose your number were 1 and 2 and 3.
supposed you wanted to know how many ways you could get the numbers when the 2 and the 3 could not be together.
assume they had to be together.
you then get 2! ways that can happen.
you then multiply that by 2 to get 4 ways that can happen.
assign A and B to 2 and 3 and you get:
1AB
1BA
AB1
BA1
that's the 4 ways they can be together.
3! = 6
6 - 4 = 2
that mans 2 ways that cannot be together.
those ways are:
A1B
B1A
replace A with 2 and B with 3 and you get the answer in the format it was presented.
I believe you answer is 12.
Answer by ikleyn(53763)  (Show Source):
You can put this solution on YOUR website! .
Different positive four-digit integers are to be formed by using each of the digits 1,2,3,4 just once
in each integer. How many different such integers can be formed if the digits 3 and 4 must NEVER be next to each other?
A. 4
B. 8
C. 12
D. 16
E. 24
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The idea is to calculate the number of all possible permutations of the four given digits
and then subtract all unfavorable arrangements, where 3 and 4 are next to each other.
The number of all possible permutations of the four given digits is 4! = 1*2*3*4 = 24.
The unfavorable arrangements are of two kinds.
One kind of unfavorable arrangements is when adjacent 3 and 4 go in this order, '34'.
Then we consider this block as one entity X, and we have 3 (three) objects to permutate:
the digits 1, 2 and the block X.
It gives 3! = 1*2*3 = 6 four-digit numbers with the block '34'.
The other kind of such arrangements is when adjacent 3 and 4 go in this order, '43'.
Then we consider this block as one entity Y, and we have 3 objects to permutate:
the digits 1, 2 and the block Y.
It gives 3! = 1*2*3 = 6 four-digit numbers with the block '43'.
Thus the number of favorable arrangements is 4! - 2*3! = 24 - 2*6 = 24 - 12 = 12.
ANSWER. The number of favorable arrangements is 12.
Solved.
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