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Question

In how many ways from 1 to 100 ( both inclusive) can we choose three numbers, that are no consecutive.

VD

mikemcgarry · Answer

Hi, there. I'm happy to help with this.

This is a really challenging question. First of all, I want to point out that this is way beyond what you would be expected to do on the GMAT. It's way beyond any of the combinatorics they ask on the AP Statistics exam. It's a hard question. I played around with it, and from what I can tell, there's no elegant pencil-and-paper type solution. If someone can propose one, I would love to know it. I solved this with technology. I'll explain my procedure. I used a wonderful statistics/modelling program called Fathom, which I highly recommend. On a spreadsheet in Fathom, I did the following: (a) column #1 = the possible values for the first number, the minimum of the three -- this was simply a list from 1 to 96 (the largest possible minimum is 96, in the trio 96-98-100; any higher, and two+ of the numbers would have to be consecutive) (c) column #2 = the allowed range for the last two --- for any given minimum, the very next number cannot be in the range. For example, if 24 is the minimum, then 25 cannot be a number in one of the allowed trios. For minimum = 24, the allowed range for the last two would be from 26 to 100, a range of 75 values. column #2 = 99 - (column #1) (d) column #3 = the total number of pairs possible in that range (includes both allowed and not-allowed pairs). When minimum = 24, there are 75 values from which to choose for the final two, and the number of pairs one could pick is 75-C-2 (that is, "75 choose 2"). This column is calculated as: column #3 = (column #2)-C-2 = / , where "!" is the factorial symbol. (e) column #4 = the number of excluded pairs in that range. For example, when minimum = 24, we will pick the remaining two from the range 26 to 100, so we are guaranteed that neither will be consecutive with 24, but the two in this pair could still be consecutive with each other. The possible consecutive pairs would be (26, 27), (27, 28), . . . . (99, 100). The starting value of one of these consecutive pairs could be any value in the range except for 100, so the number of pairs of consecutive numbers in this range would be column #4 = (column #2) - 1 (f) column #5 = the number of permitted pairs, i.e. the number of non-consecutive pairs for the last two numbers. Here, column #3 is the total number of pairs in that range, and column #4 is the number that are not permitted, so I simply subtracted those. column #5 = (column #3) - (column #4) After all that, I simply too the sum of column #5, adding up the allowed cases for each different value of the minimum number, to get the total number of allowed cases --- that is, the total number of sets of three numbers, all numbers from 1 to 100, and no two of the number consecutive. The number I got was 152096 . The total number of three-number combinations you could pick from the 100 possibilities is of course 100-C-3 = / =161700 Therefore, if three numbers from 1 to 100 were selected, without replacement, the probability that none of the three would be consecutive is P = 152096/161700 = 0.94060606 = 94% chance Please let me know if you have any questions on what I've said here. Mike :-)

KarishmaB · Answer

If the question implies that no 2 numbers of the 3 chosen should be consecutive, then this is how I would do it:

Total number of ways of choosing 3 numbers out of 100 = 100C3 = 100*99*98/3!

Number of selections in which all 3 numbers will be consecutive = 98 (the 3 numbers chosen would be (1, 2, 3) or (2, 3, 4) and so on till (98, 99, 100)

Number of selections in which only 2 numbers will be consecutive: There will be 2 cases
Case 1:
Choose any one pair from the following 97 pairs: (2, 3), (3, 4) ... (98, 99). Then select another number in 96 ways (say you select (2, 3). Now you cannot select 1, 2, 3 and 4 anymore so you have 96 numbers for the last spot)
Number of possible selections = 97*96
Case 2:
Choose any one pair of the two given pairs: (1, 2) or (99, 100). Then select another number in 97 ways (say you select (1, 2). Now you cannot select 1, 2 and 3 anymore so you have 97 numbers for the last spot)
Number of possible selections = 2*97

Number of ways of selecting 3 numbers such that there are no consecutive numbers = 100C3 - 98 - (97*96) - (2*97) = 152096

The calculations are a little crazy here.

KarishmaB · Answer

I think I have made the solution more complicated than it needs to be. Thinking back on the solution, I realized that there is a simpler method. (though it didn't occur to me when I first looked at the question)

We need to find the number of triplets in which there is at least one pair of consecutive numbers. 
Out of the possible 99 pairs (1,2), (2, 3), (3, 4)....(99, 100), choose any one in 99 ways.
Then choose the third number in 98 ways (anything except the chosen pair)
So, 99*98 gives us triplets where all 3 numbers are consecutive as well as those where only 2 are consecutive. There is one catch though. The triplets where all 3 numbers are consecutive have been double counted. Think: When you picked (2, 3) as the pair and then 4 as the third number, you got (2, 3, 4). When you picked (3, 4) as the pair and 2 as the third number, you again got (2, 3, 4). Hence you need to the subtract the number of ways in which you get all three numbers consecutive. There are 98 such ways (1, 2, 3), (2, 3, 4)....(98, 99, 100) etc
Number of ways of selecting 3 numbers such that there are no consecutive numbers = 100C3 - (99*98 - 98) = 152096

LalaB · Answer

consecutive integers are x;+1;x+2
the last consecutive numbers are 98 99 100
to get all non consecutive integers between 1 and 100 inclusive, we should choose any 98 numbers , then 97 , then 96

so, the answ is 98*97*96/3! =152096

LalaB · Answer

2nd approach is

100*98*96/3! - 98*96/2!=152096

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