In how many ways committee of 4 people can be chosen if there are 10 men and 6 women and we require that we need 3 men and 1 woman be in committee?
Well, let's break it down. To choose 3 men from a group of 10, you can think of it as picking 3 out of 10 - which can be represented as a combination. Using the "choose" notation, this is written as 10C3, which is equal to 10! / (3! * (10-3)!) = 120.
Similarly, to choose 1 woman from a group of 6, it's 6C1 = 6.
Since we need both 3 men and 1 woman, we need to multiply those numbers together. So, 120 * 6 = 720.
That means there are 720 ways to form a committee with 3 men and 1 woman from a pool of 10 men and 6 women. Let's just hope they all get along and don't start a game of "he said, she said" during their meetings!
To find the number of ways to choose a committee of 4 people, with 3 men and 1 woman, from a pool of 10 men and 6 women, we can use combination formula.
First, let's calculate the number of ways to choose 3 men from 10 men. This can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!)
Where n is the total number of items and r is the number of items to be chosen.
C(10, 3) = 10! / (3!(10-3)!)
= 10! / (3!7!)
= (10 * 9 * 8) / (3 * 2 * 1)
= 120
Next, let's calculate the number of ways to choose 1 woman from 6 women:
C(6, 1) = 6! / (1!(6-1)!)
= 6! / (1!5!)
= 6
Finally, we need to multiply these two values to get the total number of ways to choose the committee:
Total number of ways = C(10, 3) * C(6, 1)
= 120 * 6
= 720
So, there are 720 ways to choose a committee of 4 people, with 3 men and 1 woman, from a group of 10 men and 6 women.
To solve this problem, we can use the concept of combinations.
First, let's calculate the number of ways to choose 3 men out of the 10 available. This can be calculated using the combination formula, also known as "n choose r". The formula for calculating combinations is:
C(n, r) = n! / (r! * (n - r)!)
Where "n" represents the total number of options, and "r" represents the number of options to be chosen.
So, for choosing 3 men out of 10, we have:
C(10, 3) = 10! / (3! * (10 - 3)!)
Simplifying this expression:
C(10, 3) = 10! / (3! * 7!)
Now, let's calculate the number of ways to choose 1 woman out of the 6 available:
C(6, 1) = 6! / (1! * (6 - 1)!)
Simplifying this expression:
C(6, 1) = 6! / (1! * 5!)
Now, since we need to choose 3 men and 1 woman, we multiply the two combinations together:
C(10, 3) * C(6, 1) = (10! / (3! * 7!)) * (6! / (1! * 5!))
Simplifying further:
C(10, 3) * C(6, 1) = (10 * 9 * 8) / (3 * 2 * 1) * 6
After simplification, we get:
C(10, 3) * C(6, 1) = 10 * 3 * 8 * 6 = 1440
Therefore, there are 1440 ways to choose a committee of 4 people, with 3 men and 1 woman, from a group of 10 men and 6 women.
Answer
3 men from the 10 men ----> C(10,3)
1 woman from the 6 women ---> C(6,1)
So, what do you think?