In a particular year, a company awards scholarships to 4 applicants from a group of $n$ applicants. If the order of selection is not taken into account, the number of ways in which the 4 scholarship recipients can be chosen is 7315.

1. Relate the number of combinations to the number of permutations and evaluate

$\text{Number of ways} = 7315 \times 4! = 175\,560$

1. Formulate an equation for the number of ways to choose 4 recipients from $n$ applicants.

$\binom{n}{4} = 7315$

2. Solve for $n$ by testing integer values or using a graphing calculator.

Estimating $n$ since $n^4 \approx 7315 \times 24 = 175560$, we expect $n \approx \sqrt[4]{175560} \approx 20.4$.
Testing integer values:
When $n = 21$, $\binom{21}{4} = 5985$
When $n = 22$, $\binom{22}{4} = 7315$

Hence, $n = 22$.

1. Find the number of ways to position a contiguous block of 4 seats within the two rows.

The 4 scholarship recipients must sit together, so they must occupy 4 consecutive seats in the same row.
Number of valid blocks of 4 seats in the top row (6 seats) $= 6 - 4 + 1 = 3$
Number of valid blocks of 4 seats in the bottom row (7 seats) $= 7 - 4 + 1 = 4$
Total valid positions for the block $= 3 + 4 = 7$

2. Formulate the total number of seating arrangements.

For each valid block position, the 4 scholarship recipients can be arranged among themselves in $4!$ ways.
The remaining 9 applicants can be arranged in the remaining 9 seats in $9!$ ways.
Total number of seating arrangements $= 7 \times 4! \times 9!$

3. Calculate the final value.

$= 7 \times 24 \times 362\,880$
$= 60\,963\,840$