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let x be number 24
x135
can be arranged in 24 ways
and 42 now be y
y135 can be arranged in 24 ways
total ways to arrange 5 digits is 120
120- ( 24+24) ; 72
option C

jackhp
A 5-digit identification code is to be created as a sequence that contains each integer in the set {1, 2, 3, 4, 5} exactly once. If all identification codes are possible except those containing even integers next to each other, how many different identification codes are possible?

A. 120
B. 96
C. 72
D. 48
E. 24
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Is there another approach to this question? How would we solve it using the Permutation formula?­
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­A 5-digit identification code is to be created as a sequence that contains each integer in the set {1, 2, 3, 4, 5} exactly once. If all identification codes are possible except those containing even integers next to each other, how many different identification codes are possible?

We must create codes such that even integers are separated by odd integers.

First arrange the odd integers. There are 3 odd integers. So, they can be arranged in 3! = 6 ways.

Now, having arranged the odd integers, we have the following possible slots for the 2 even integers:

_ O _ O _ O _

We see that there are 4 slots we can choose for the even integers.

So, for example, we can choose the following:

2 O 4 O _O _

Thus, we are choosing 2 out of 4 slots. So, the ways to chose the slots are 4C2 = 6.

There are 2 ways to order the 2 even integers in the 2 slots they occupy. So, we have 6 × 2 = 12 ways to arrange the even integers.

So, with 6 ways to arrange the odd integers and 12 ways to arrange the even integers, we have 6 × 12 = 72 total posssible arrangements. 

A. 120
B. 96
C. 72
D. 48
E. 24


Correct answer: C
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­Let's just do Total minus evens next to each other and call it a day:

­
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­Let's just do Total minus evens next to each other and call it a day:

­
­Can you explain how you got the 4 * 3 * 2 * 2
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Hi chetan2u,

This makes sense and thank you for the explanation. I just wanted to test my understanding.

If for example the question was {1,2,3,4,5,6} for a 6-digit code - would the answer be 6! * (4! * 3!) - because there are 6 ways to arrange the 3 even digits?

Thanks in advance!
chetan2u
jackhp
A 5-digit identification code is to be created as a sequence that contains each integer in the set {1, 2, 3, 4, 5} exactly once. If all identification codes are possible except those containing even integers next to each other, how many different identification codes are possible?

A. 120
B. 96
C. 72
D. 48
E. 24


Total ways to create 5-digit identification code = 5!

Restriction: Both even number cannot be together.

When both the even digits are taken together as one digit, we have 4 digits, that is 1, 3, 5 and (24) => 4! to arrange 4 digits and 2 & 4 can be arranged in 2! ways within themselves. Hence, 4!*2

Ways we are looking at = 5!-4*2! = 120-48 = 72


C
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