Each digit 1 through 5 is used exactly once to create a 5-di

Such questions are best solved using the complement approach. You need to find arrangements in which 3 and 4 are not adjacent. Instead, work on finding the arrangements in which they are together since it is much easier to do.

Number of arrangements using 5 distinct digits = 5!
Number of arrangements in which 3 and 4 are adjacent - consider 3 and 4 together as one group. Now you have 4 numbers/groups to arrange which can be done in 4! ways. In each of these arrangements, 3 and 4 can be arranged as 34 or 43. So we need to multiply 4! by 2 to give all arrangements where 3 and 4 are adjacent to each other e.g. 34251, 43251, 23415, 32415 ...
Number of arrangements in which 3 and 4 are not adjacent = 5! - 2*4! = 72
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Total number of 5 digit numbers that can be formed = 5! = 120.
Total number of 5 digit numbers in which 3 and 4 are adjacent = 48. This can be calculated by taking 3 and 4 as a pair and considering positions where they can appear.
Number of 5 digit numbers in which 3 and 4 are not adjacent = Total number of 5 digit numbers - Total number of 5 digit numbers in which 3 and 4 are adjacent.
Ans = 120 - 48 = 72.

Total number of arrangements of 5 digit integer = 5! = 120

Let 3 and 4 be single digit like { 1 2 X 5 }
This set arrangement is 4! = 24
As 3 and 4 can be interchanged between them 2(24) = 48

So 120-48 =72
Pls correct me if my solution is wrong!

combinatorics and permutations eish..how do you guys make out on which way to use,whether combination or permutation,when you have such problem to solve?..when do you use permutation and when do you use combination?

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Video on Permutations: https://youtu.be/LFnLKx06EMU
Video on Combinations: https://youtu.be/tUPJhcUxllQ
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VeritasPrepKarishma can you please explain how you draw the conclusion that there are 4 numbers/groups to arrange? There are only 2 numbers and the second part of your answer addresses switching their order. I am not understanding this conclusion. Thank you!

You need to create a 5 digit integer using digits from 1 to 5. You will do that in various ways such as 12345, 23145, 45231 etc

If you club 3 and 4 together, you get a group {34}.

Now you have 3 leftover digits: 1, 2 and 5.
You have arrange 1, 2, 5 and {34}. So you have to arrange 4 digits/groups.
You can do this in various ways such as 12534, 21345 etc...

Since instead of {34}, you can have {43} too, you multiply by 2.
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