A party of 6 is to be formed from 10 boys and 7 girls so as to include 3 boys and 3 girls. In how many different ways can the party be formed if two particular girls refuse to join the same party?

To form a party of 6 with 3 boys and 3 girls from 10 boys and 7 girls, we first calculate the total combinations without restrictions. The number of ways to choose 3 boys from 10 is ( \binom{10}{3} ), and the number of ways to choose 3 girls from 7 is ( \binom{7}{3} ). However, since two particular girls cannot be in the same party, we need to exclude the scenariOS where both are chosen.

The total combinations without restrictions is ( \binom{10}{3} \times \binom{7}{3} ). Then, we subtract the cases where both girls are included, which is ( \binom{10}{3} \times \binom{5}{1} ) (choosing 1 more girl from the remaining 5 girls). The final count is:

[ \text{Total ways} = \binom{10}{3} \times \binom{7}{3} - \binom{10}{3} \times \binom{5}{1} ]

Calculating these values gives:

[ \binom{10}{3} = 120, \quad \binom{7}{3} = 35, \quad \binom{5}{1} = 5 ]

Thus:

[ \text{Total ways} = 120 \times 35 - 120 \times 5 = 4200 - 600 = 3600 ]

Therefore, there are 3600 different ways to form the party under the given conditions.