# Tuples with 4 consecutive integers from 1 to 50

How many ways are there to pick an ordered tuple of 5 elements from the set of positive integers between 1 and 50 inclusive, provided that there must be at least one consecutive sequence of 4 consecutive elements?

So what I think is somehow we need to use inclusion exclusion principle ; So I proceed by calculating sequence with 5 consecutive element but then I have no idea how to proceed with this?

• You have ruined your own question with that edit. It is now completely unintelligible! – TonyK Oct 17 '19 at 11:46
• @TonyK Oh sorry for that. – maths student Oct 17 '19 at 11:51

There are 47 sets of 4 consecutive elements, from $$\{1,2,3,4\}$$ to $$\{47,48,49,50\}.$$ For each of these, we can choose a fifth element for our 5-tuple in 46 ways, and this fifth element can be placed in the first or last position in the 5-tuple. So far our count is

$$47 \cdot 46 \cdot 2$$

Notice this count would include the 5-tuples with five consecutive elements as well; however each of these 5-tuples with 5 consecutive elements would have been counted twice, either by choosing as our fifth element the first or the last of the five consecutive numbers. [e.g the 5-tuple $$(23,24,25,26,27)$$ could be created by placing $$23$$ in front of the 4-tuple $$(24,25,26,27)$$ or by placing $$27$$ at the end of the 4-tuple $$(23,24,25,26)$$.]

Thus we need to subtract the number of 5-tuples with 5 consecutive elements, of which there are clearly 46. So our final count will be

$$47 \cdot 46 \cdot 2 - 46 = \boxed{4278}$$

N.B. I've assumed in the above that the 5-tuples are to be made from five different elements; if repetition were allowed, then in the first part of the computation, when choosing the fifth element there would be 50 choices rather than 46, so then the count would be

$$47 \cdot 50 \cdot 2 - 46 = \boxed{4654}$$

If I understand you well, you have

i) 1 group $$[1 \ 2\ 3\ 4]$$, with 45 other possible non consecutive elements, located at 2 possible positions (beggining or end),

ii) 1 group $$[47 \ 48\ 49\ 50]$$, with 45 other possible non consecutive elements, located at 2 possible positions,

iii) 45 other groups of 4 consecutive elements, with 44 other possible non consecutive elements, located at 2 possible positions,

iv) the 1 group in i) of 4 consecutive elements, with 1 next possible consecutive elements $$[5]$$, located at the beginning,

iv) the 1 group in ii) of 4 consecutive elements, with 1 next possible consecutive elements $$[46]$$, located at the end,

iv) the 45 groups in iii) of 4 consecutive elements, with 2 possible consecutive elements, located at the non consecutive position (begginning or end accordingly),

v) 46 groups of 5 consecutive elements.

In total, $$1\cdot45\cdot2+1\cdot45\cdot2+1+1+45\cdot2+45\cdot44\cdot2+46=4278$$ possible 5 tuples.

• You are missing the 46 groups with 5 consecutive elements: $[1,2,3,4,5],\ldots,[46,47,48,49,50]$. At least by my understanding of the problem, they should be valid solutions. – Ingix Oct 17 '19 at 8:20