组成委员会

先选 3 人组委会：$\binom{10}{3}$。

剩下 7 人中再选 4 人活动委员会：$\binom{7}{4}$。

因此总数：

$$\binom{10}{3}\binom{7}{4}=120\times 35=4200.$$

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Original Explanation

To solve this problem, we consider:

1. Choosing the 3-member organizing committee:
   The number of ways to choose 3 members out of 10 is given by the combination: $$\binom{10}{3}$$

2. Choosing the 4-member event committee:
   After selecting the 3 members for the organizing committee, 7 members remain. The number of ways to choose 4 members from these 7 is: $$\binom{7}{4}$$

3. Multiplying both choices:
   Since the committees are distinct (i.e., someone chosen for one cannot serve on the other), the total number of ways is the product of the two: $$\binom{10}{3} \times \binom{7}{4}$$

So the number of ways the club can form the two committees is: $$\binom{10}{3} \times \binom{7}{4} = 120 \times 35 = 4200$$