How many ways can the integers 1,1,2,3,4,5,6 be arranged in a row, so that no integer is immediately adjacent to two strictly larger integers?
NOTE: The sequence 3,1,2,4,5,6,1 is not valid as 1 is adjacent to 2 and 3, both of which are strictly larger than 1.
The sequence 3,1,1,2,4,5,6, satisfies the conditions of the question.
To find the number of ways the integers can be arranged, we can use the concept of permutations.
We have a total of 7 integers: 1, 1, 2, 3, 4, 5, and 6. Since there are two 1s, we need to consider them as distinct elements. So effectively, we have 1', 1, 2, 3, 4, 5, and 6.
First, let's consider the positions of the 1s. We can arrange them in 3 ways: 1'-1, 1-1', 1'-1'.
Next, we have 5 remaining integers to arrange: 2, 3, 4, 5, and 6.
Now, we need to make sure that no integer is immediately adjacent to two strictly larger integers. To satisfy this condition, we can place the integers following the pattern:
Larger-Int1-Larger-Int2
Int1 can be 3, 4, or 5, and Int2 can be 4, 5, or 6:
For Int1 = 3:
- Int2 can be 4, 5, or 6, so we have 3 choices.
For Int1 = 4:
- Int2 can be 5 or 6, so we have 2 choices.
For Int1 = 5:
- Int2 can only be 6, so we have 1 choice.
Multiplying the choices for Int1 and Int2, we get a total of 3 * 3 + 3 * 2 + 1 * 1 = 14 ways to arrange the remaining integers.
Finally, multiplying the total number of ways to arrange the 1s (3) by the total number of ways to arrange the remaining integers (14), we get the answer: 3 * 14 = 42.
Therefore, there are 42 ways to arrange the integers 1, 1, 2, 3, 4, 5, 6 in a row such that no integer is immediately adjacent to two strictly larger integers.