anybody please solve this math problem for me

[T.Q]

help me ,solve this math problem with explaintion how to solve it,aslo show your working please


6 (a) Calculate the number of different 6-digit numbers which can be formed using the digits
0, 1, 2, 3, 4, 5 without repetition and assuming that a number cannot begin with 0. [2]

(b) A committee of 4 people is to be chosen from 4 women and 5 men. The committee must contain at
least 1 woman. Calculate the number of different committees that can be formed. [4]

[astarmathsandphysics]

help me ,solve this math problem with explaintion how to solve it,aslo show your working please


6 (a) Calculate the number of different 6-digit numbers which can be formed using the digits
0, 1, 2, 3, 4, 5 without repetition and assuming that a number cannot begin with 0. [2]

(b) A committee of 4 people is to be chosen from 4 women and 5 men. The committee must contain at
least 1 woman. Calculate the number of different committees that can be formed. [4]

a)If it cant begin with 0 then you have a choice of 5 for the 1st number, and the second can be any of the remaining 5, then the third any of the remaining 4 etc ANS=5*5*4*3*2*1=600
b)At least one woman
Suppose one woman and three men. We have a choice of 4 women and the men can be picked in 5C3=10 ways so 4*10
=40 ways here
Suppose two women and two men. We can choose 2 women in 4C2=6 and the men in 5C2=10 ways so6*10=60 ways here
Suppose 3 women and 1 man. We can choose the women in 4C3=4 ways and the man in 5C1=5 ways so 20 ways here
so 40+60+20=110 ways altogether

[T.Q]

help me ,solve this math problem with explaintion how to solve it,aslo show your working please


6 (a) Calculate the number of different 6-digit numbers which can be formed using the digits
0, 1, 2, 3, 4, 5 without repetition and assuming that a number cannot begin with 0. [2]

(b) A committee of 4 people is to be chosen from 4 women and 5 men. The committee must contain at
least 1 woman. Calculate the number of different committees that can be formed. [4]

a)If it cant begin with 0 then you have a choice of 5 for the 1st number, and the second can be any of the remaining 5, then the third any of the remaining 4 etc ANS=5*5*4*3*2*1=600
b)At least one woman
Suppose one woman and three men. We have a choice of 4 women and the men can be picked in 5C3=10 ways so 4*10
=40 ways here
Suppose two women and two men. We can choose 2 women in 4C2=6 and the men in 5C2=10 ways so6*10=60 ways here
Suppose 3 women and 1 man. We can choose the women in 4C3=4 ways and the man in 5C1=5 ways so 20 ways here
so 40+60+20=110 ways altogether

im very thankful
really appreciated