counting problem

[steven_ghoos]

Can someone explain to me the following calculation?

How many three-digit numbers are there such that all three digits are different and the first digit is not zero?

Thanks

[Anurag@Gurome]

How many three-digit numbers are there such that all three digits are different and the first digit is not zero?

Say the digit is ABC.

A can be any of the 9 digits 1 - 9 ----> 9 possibility
B can be any of the 10 digits 0- 9 except the one used in A ---> 9 possibilities
C can be any of the 10 digits 0- 9 except the two used in A and B ---> 8 possibilities

Thus, total number of such numbers = 9*9*8 = 648

[steven_ghoos]

Thanks, it's clear now.

Below is another I'm coping with...

From a total of 5 boys and 4 girls, how many 4-person committees can be selected if the committee must have exactly 2 boys and 2 girls?

How do you come to 5!/2!2! * 4!/2!2! and how do you come to 10*6 -> calculate or can you use a method?

Regards

[GMATGuruNY]

Thanks, it's clear now.

Below is another I'm coping with...

From a total of 5 boys and 4 girls, how many 4-person committees can be selected if the committee must have exactly 2 boys and 2 girls?

How do you come to 5!/2!2! * 4!/2!2! and how do you come to 10*6 -> calculate or can you use a method?

Regards

Number of ways to combine 2 boys from 5 choices = 5C2 = (5*4)/(2*1) = 10.
Number of ways to combine 2 girls from 4 choices = 4C2 = (4*3)/(2*1) = 6.
To combine our number of choices from the boys with our number of choices from the girls, we multiply:
10*6 = 60.

[steven_ghoos]

Thanks, it's clear now.

Below is another I'm coping with...

From a total of 5 boys and 4 girls, how many 4-person committees can be selected if the committee must have exactly 2 boys and 2 girls?

How do you come to 5!/2!2! * 4!/2!2! and how do you come to 10*6 -> calculate or can you use a method?

Regards

Number of ways to combine 2 boys from 5 choices = 5C2 = (5*4)/(2*1) = 10.
Number of ways to combine 2 girls from 4 choices = 4C2 = (4*3)/(2*1) = 6.
To combine our number of choices from the boys with our number of choices from the girls, we multiply:
10*6 = 60.

Sorry, but this is not clear to me. Can you please explain in more detail?

[Anurag@Gurome]

From a total of 5 boys and 4 girls, how many 4-person committees can be selected if the committee must have exactly 2 boys and 2 girls?

Number of ways to select exactly 2 boys and 2 girls = (Number of ways to select 2 boys out of 5)*(Number of ways to select 2 girls out of 4) = (5C2)*(4C2) = [5!/(3!*2!)]*[4!/(2!*2!)] = 10*6 = 60

How do you come to 5!/2!2! * 4!/2!2! and how do you come to 10*6...

The red part is wrong. It should be 5!/2!*3! as mentioned earlier.

[GMATGuruNY]

Thanks, it's clear now.

Below is another I'm coping with...

From a total of 5 boys and 4 girls, how many 4-person committees can be selected if the committee must have exactly 2 boys and 2 girls?

How do you come to 5!/2!2! * 4!/2!2! and how do you come to 10*6 -> calculate or can you use a method?

Regards

Number of ways to combine 2 boys from 5 choices = 5C2 = (5*4)/(2*1) = 10.
Number of ways to combine 2 girls from 4 choices = 4C2 = (4*3)/(2*1) = 6.
To combine our number of choices from the boys with our number of choices from the girls, we multiply:
10*6 = 60.

Sorry, but this is not clear to me. Can you please explain in more detail?

Below is a description of the method I use to count combinations.

What is the number of different committees of 4 people that can be selected from a group of 10 people?

I use what some call the "slot method":

Draw a slot for each choice that is being made. Since we want 4 people on the committee, we draw 4 slots:

__ * __ * __ * __

We have 10 choices for the 1st slot (because we have 10 people to choose from).
We have 9 choices for the 2nd slot (because we used 1 person to fill slot 1, leaving us 10-1=9 choices for the 2nd slot).
We have 8 choices for the 3nd slot (because we used 2 people to fill slots 1 and 2, leaving us 10-2=8 choices for the 3rd slot).
We have 7 choices for the last slot (because we used 3 people to fill slots 1, 2 and 3, leaving us 10-3=7 choices for the last slot).

So far we have: 10 * 9 * 8 * 7

If we multiply the numbers above, we'll be counting the number of ways we can arrange 4 people from the 10 choices. For example, ABCD is a different arrangement from BCAD. The mathy word for this is permutation. There are 10*9*8*7 = 5040 ways to arrange 4 people from 10 choices.

But the problem above is asking how many groups can be made, and ABCD and BCAD are the same group (the mathy word for this is combination). So we don't overcount the duplicate combinations, we need to divide by (number of slots)!. This division will remove from our result all the duplicate combinations (so we don't overcount ABCD and BCAD as different combinations).

Since in this problem we have 4 slots, we divide by 4!:

10*9*8*7/4*3*2*1 = 210. So there are 210 ways to combine 4 people from 10 choices. Notice that the number of possible combinations is smaller than the number of possible permutations.

[steven_ghoos]

From a total of 5 boys and 4 girls, how many 4-person committees can be selected if the committee must have exactly 2 boys and 2 girls?

Number of ways to select exactly 2 boys and 2 girls = (Number of ways to select 2 boys out of 5)*(Number of ways to select 2 girls out of 4) = (5C2)*(4C2) = [5!/(3!*2!)]*[4!/(2!*2!)] = 10*6 = 60

How do you come to 5!/2!2! * 4!/2!2! and how do you come to 10*6...

The red part is wrong. It should be 5!/2!*3! as mentioned earlier.

thank you, but I still don't see how you get there? can you explain? and do you make the calculation to come to 10*6 or is there a formula that simplifies this?

Thanks a lot!

[Anurag@Gurome]

thank you, but I still don't see how you get there? can you explain? and do you make the calculation to come to 10*6 or is there a formula that simplifies this?

Thanks a lot!

Yes, I have used the standard combination formula nCr = n!/[(n - r)!*r!], which gives the number of possible different selections of r objects out of a total n different objects.

Any standard book on permutation-combination should have a section which gives detail explanation of this formula. You can also check here : https://www.mathsisfun.com/combinatorics ... tions.html

[steven_ghoos]

Nice. Thank you all.

So I take 2 slots out of 5 in the first group (5*4/2!) and multiply it with the 2 slots out of the second group (4*3/2!).

Or, using the general rule for combinations: [5!/(5-2)!*2!]*[4!/(4-2)!*2!]

And if it would be permutations rather than combinations, it would be 5*4*4*3 = 240 (no removal of duplications)