Question

Let the set $A = \{1, 2, 3, 4, 5, 6\}$ and $B = \{a, b, c\}$. The elements of $A \cup B$ are arranged in such a way that no two letters occur consecutively. If the number of all possible ways in which this can be done is $\alpha$, then the value of $\frac{\alpha}{6!}$ is ___.

Answer 1

210

Answer 2

To solve this problem, we first arrange the 6 distinct numbers from set A. This can be done in $6!$ ways. These 6 numbers create $6+1=7$ possible gaps where the 3 distinct letters from set B can be placed. To ensure that no two letters occur consecutively, each of the 3 letters must be placed in a different gap. The number of ways to choose 3 gaps from 7 and arrange the 3 distinct letters in them is given by the permutation formula $P(7, 3)$.

$P(7, 3) = \frac{7!}{(7-3)!} = \frac{7!}{4!} = 7 \times 6 \times 5 = 210$.

The total number of possible arrangements, $\alpha$, is the product of the ways to arrange the numbers and the ways to place the letters: $\alpha = 6! \times P(7, 3) = 6! \times 210$.

The question asks for the value of $\frac{\alpha}{6!}$: $\frac{\alpha}{6!} = \frac{6! \times 210}{6!} = 210$.