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­To calculate total number of ways in which atleast one is even means we can find its opposite and subtract it from total (number of ways 2 numbers can be selected from given set of 5 numbers )
Total number of wasy (2 numbers selected from 5 numbers) = 5C2 = 10 
No. of ways both are odd (i.e numbers to be chosen from 1,3 & 5 ) = 3C2 = 3
Hence to find atleast one is even = 10-3 = 7­
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We should have at least 1 even integer. Consider 2 cases -

Case 1 - If the even integer we have is 2, we can select the second number in 4 ways (out of 1,3,4 and 5).

Case 2 - If the even integer we have is 4, we can select the second number in 4 ways (out of 1,2,3 and 5).

However, in both the cases, the selection of numbers (2,4) must be included. Since the question only asks about selection and not arrangement, we will have to remove the duplicate case.

Thus, the final answer would be 4 ways + 4 ways - 1 way = 7 ways.

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A_Nishith
Two numbers selected are within the given set of (odd, even) combination : { (O,O),(E,O),(O,E),(E,E) }
Ask is at least one Even is selected i.e Total possibility - No Even.
Given set {1,2,3,4,5}, Even {2,4} , Odd {1,3,5}
Total possible ways = 5*4 = 20
No Even i.e both Odd = 3*2 = 6
Therefore, Required answer must be 20-6 = 14

BUT answer is 14/2=7 ---- HOW

Because order doesn't matter.
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Answer is Option C - 7

Here, we are just selecting 2 numbers, so the order of the selection doesn't matter.

We can find the total number of ways to select at least 1 even integer by:
(Total number of ways to select 2 numbers from 5 numbers) - (Number of ways to select only odd integers)

5C2 - 3C2 = 10 - 3 = 7.
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(Total combos) - (odd only) is the most efficient method to solve this problem.

The solution below approaches the problem from the (even only) + (even + odd perspective). Keep in mind that order does not matter, as the problem does not contain language specifying that case.

Even only has only one combination. 2c2=1
Even + Odd is 3C1 * 2C1 = 6
The total number of combinations are 7.
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