Question:

Step 1: Determine the number of ways to rank 10 songs such that there are exactly 3 Indian classic songs in the top 5 positions. There are 6 Indian classic songs and 4 Western songs. We need to form a ranking of 10 songs. 1. **Select Indian songs for the top 5:** Choose 3 Indian classic songs from the 6 available. This can be done in $\binom{6}{3}$ ways. $$ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 $$ 2. **Select Western songs for the top 5:** Since there must be exactly 3 Indian songs in the top 5, the remaining $5-3=2$ positions in the top 5 must be filled by Western songs. Choose 2 Western songs from the 4 available. This can be done in $\binom{4}{2}$ ways. $$ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 $$ 3. **Arrange the songs in the top 5 positions:** The 5 selected songs (3 Indian, 2 Western) can be arranged in the top 5 positions in $5!$ ways. $$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $$ 4. **Arrange the remaining songs in the bottom 5 positions:** After selecting 3 Indian and 2 Western songs for the top 5, there are $6-3=3$ Indian songs and $4-2=2$ Western songs remaining. These $3+2=5$ songs must fill the bottom 5 positions (ranks 6 through 10). They can be arranged in $5!$ ways. $$ 5! = 120 $$ The total number of ways to rank the songs such that there are exactly 3 Indian classic songs in the top 5 is the product of the ways for each step: $$ \text{Total ways} = \binom{6}{3} \times \binom{4}{2} \times 5! \times 5! $$ $$ \text{Total ways} = 20 \times 6 \times 5! \times 5! $$ $$ \text{Total ways} = 120 \times 5! \times 5! $$ Since $120 = 5!$, we can substitute this value: $$ \text{Total ways} = 5! \times 5! \times 5! = (5!)^3 $$ Thus, the number of ways of ranking so that there are exactly 3 Indian classic songs in the top 5 is $(5!)^3$.