Choose 6 distinct numbers from set of numbers so that at least 2 of them are consecutive

combinatorics

Find the number of ways in which from the set 1, 2, …, 49 you can choose six
different numbers so that at least two of them are consecutive.

I tried to solve it in this way:
Two consecutive numbers can be:

  • 1,2 | 1,3 | 1,4 | … | 1,49
  • 2,3 | 2,4 | 2,5 | … | 2,49

and so on.So total ways of picking two consecutive numbers will be 1+2+3+..+48 which is 1176. And number of ways of picking other 4 numbers will be 47 * 46 * 45 * 44. So total combination will be 1176 * 47 * 46 * 45 * 44. Am I right? (sorry for bad english)

Best Answer

The number of ways to choose 6 numbers will be $\binom{49}{6}$ and the number of ways to choose 6 non-consecutive numbers will be $\binom{44}{6}$ (refer to this).

Hence, the answer is $\binom{49}{6} - \binom{44}{6} = \boxed{6924764}$.

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