froma group of 6 ladies and 4 men, determine in how many ways a committee of 4 people cxan be selected.

a. with no restirctions
b. 4 women
c. 3 women and 1 men
d. 2 women and 2 men
e. 4 men

With no restrictions, just select any four from ten, which is:

factorial(10) / (factorial(4) x factorial(6))
= (10x9x8x7x6x5x4x3x2x1) / {(6x5x4x3x2x1) x (4x3x2x1)}
= (10x9x8x7) / (4x3x2x1).

If you've got to have four women and two men, then you're choosing four women from six, and also two men from four. The number of ways you can choose four women from six is:

(6x5x4x3x2x1) / {(4x3x2x1) x (2x1)} = (6x5) / (2x1) = 15 [Just to demonstate that, you can easily enumerate them all in this instance, since the number is very small. Call them A, B, C, D, E and F. All the possible combinations are as follows: ABCD, ABCE, ABCF, ABDE, ABDF, ABEF, ACDE, ACDF, ACEF, ADEF, BCDE, BCDF, BCEF, BDEF, CDEF. That's 15 in total.]

The number of ways to choose two men from four is (4x3x2x1) / {(2x1) x (2x1)} = 6. [Again, you can easily enumerate them all here. Call them P, Q, R and S. All the possible combinations are PQ, PR, PS, QR, QS and RS. That's 6 in total.]

The total number of ways of getting BOTH (four women from six) AND (two men from four) is just the product of those two previous answers.

Are you able to tackle the remaining parts of the question now?

To determine the number of ways a committee of 4 people can be selected from a group of 6 ladies and 4 men, we can use the concept of combinations.

a. With no restrictions:
In this case, we can select any combination of 4 people from the total group of 10. The number of ways to do this can be calculated using the formula for combinations, which is nCr = n! / (r!(n-r)!), where n is the total number of people and r is the number of people to be selected.

So, for this case, the number of ways to select a committee with no restrictions is 10C4 = 10! / (4!(10-4)!) = 210 ways.

b. With 4 women:
Since there are 6 ladies in the group, we need to select 4 of them and none of the men. The number of ways to do this is 6C4 = 6! / (4!(6-4)!) = 15 ways.

c. With 3 women and 1 man:
Here, we need to choose 3 women from the group of 6 and 1 man from the group of 4. The number of ways to do this is (6C3) * (4C1) = (6! / (3!(6-3)!) ) * (4! / (1!(4-1)!)) = 60 ways.

d. With 2 women and 2 men:
To form a committee with 2 women and 2 men, we need to select 2 women from the group of 6 and 2 men from the group of 4. The number of ways to do this is (6C2) * (4C2) = (6! / (2!(6-2)!)) * (4! / (2!(4-2)!)) = 90 ways.

e. With 4 men:
We need to select all 4 men from the group of 4, and no women. Since there are no restrictions on the selection for men, there is only 1 way to do this.

Therefore, the number of ways the committee can be selected is:
a. 210 ways
b. 15 ways
c. 60 ways
d. 90 ways
e. 1 way