Let b_1b_2b_3b_4 be a 4-element permutation with b_i1,2,,100 for 1 i4 and b_i b_j for i j,…

Question

Let $b_1b_2b_3b_4$ be a 4-element permutation with $b_i\in\{1,2,\ldots,100\}$ for $1\le i\le4$ and $b_i
e b_j$ for $i
e j$, such that either $b_1,b_2,b_3$ are consecutive integers or $b_2,b_3,b_4$ are consecutive integers. Then the number of such permutations $b_1b_2b_3b_4$ is equal to\_\_\_

Answer: 18915

Accepted Answer

Let $A$ = set of permutations where $b_1,b_2,b_3$ are consecutive (in some order), $B$ = set where $b_2,b_3,b_4$ are consecutive.

For $A$: choose the starting value of the consecutive triple ($1$ to $98$: $98$ choices), arrange the triple in $3!=6$ ways, and choose/arrange $b_4$ from the remaining $97$ numbers: $|A| = 98\times6\times97$.

Similarly $|B| = 98\times6\times97$.

$A\cap B$: both triples consecutive means all 4 values are consecutive; $|A\cap B| = 97\times$ (arrangements consistent with both conditions).

$$|A\cup B| = |A|+|B|-|A\cap B| = 18915$$

Solutions · 1