Question

Number of permutations of 1, 2, 3, 4, 5, 6, 7, 8, 9 taken all at a time, such that the digit 1 appears somewhere to the right of 2, 3 appears somewhere to the left of 4 and 5 appears in between (not necessarily exactly in middle) of 6 & 7 is $k \cdot (7!)$, then k is _____

Answer 1

6

Answer 2

Solution:

We have 9 digits and we want the permutations to satisfy three conditions:

1. 1 appears to the right of 2:
   In any permutation, the probability is $\frac{1}{2}$.

2. 3 appears to the left of 4:
   In any permutation, the probability is $\frac{1}{2}$.

3. 5 appears between 6 and 7:
   Among the three digits 5, 6, and 7, the total arrangements are $3! = 6$.
   The favorable arrangements where 5 lies between 6 and 7 are: $6,5,7$ and $7,5,6$, so there are 2 favorable cases.
   Hence, the probability is $\frac{2}{6} = \frac{1}{3}$.

Combined probability:

$$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{3} = \frac{1}{12}.$$

Since the total number of permutations of 9 digits is $9!$, the count of favorable permutations is:

$$\frac{9!}{12}.$$

We are given that this equals $k \times (7!)$. Noting that

$$9! = 9 \times 8 \times 7!,$$

we have:

$$\frac{9 \times 8 \times 7!}{12} = k \times 7!.$$

Canceling $7!$ from both sides:

$$\frac{9 \times 8}{12} = k \quad \Longrightarrow \quad \frac{72}{12} = k \quad \Longrightarrow \quad k = 6.$$

Explanation (Minimal Core):

• Requirement probabilities: $\frac{1}{2}$ (for 1 & 2), $\frac{1}{2}$ (for 3 & 4), $\frac{1}{3}$ (for 5,6,7).
• Overall probability $\frac{1}{12}$.
• Favorable count: $\frac{9!}{12} = 6 \times 7!$.
• Thus, $k = 6$.

Answer:

$k = 6$.