From a group of 7 men and 6 women, a committee of 5 people must be formed such that the committee includes at least 3 men. Determine the total number of different possible committee arrangements.

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To find the number of ways to form a 5-person committee with at least 3 men, I'll analyze the possible cases where the condition is satisfied:

### Case 1: Exactly 3 men and 2 women
- Number of ways to select 3 men from 7 men: $^7C_3 = \frac{7!}{3!(7-3)!} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$
- Number of ways to select 2 women from 6 women: $^6C_2 = \frac{6!}{2!(6-2)!} = \frac{6 	imes 5}{2 	imes 1} = 15$
- Total ways for this case: $35 	imes 15 = 525$

### Case 2: Exactly 4 men and 1 woman
- Number of ways to select 4 men from 7 men: $^7C_4 = \frac{7!}{4!(7-4)!} = \frac{7 	imes 6 	imes 5 	imes 4}{4 	imes 3 	imes 2 	imes 1} = 35$
- Number of ways to select 1 woman from 6 women: $^6C_1 = 6$
- Total ways for this case: $35 	imes 6 = 210$

### Case 3: All 5 men and 0 women
- Number of ways to select 5 men from 7 men: $^7C_5 = \frac{7!}{5!(7-5)!} = \frac{7 	imes 6}{2 	imes 1} = 21$
- Total ways for this case: $21$

### Total number of ways
$525 + 210 + 21 = 756$

Therefore, the committee can be formed in 756 different ways.

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