Problem Solving

Hi all thanks for reading

have a question on this

comp has 6 senior and 4 junior officers.

if a committee is created to be 3 senior and 1 junior officers how many diff combination's possible?

is there a formula for this type?

the answer is 80. I thought all I do for this is times 6x4=24.

You need to choose 3 senior officers from 6 and 1 junior officer from 1

So you have

6C3 = 6!/3!3! = 20

4C1 = 4!/3! = 4

then you need to multiply the these two together to get all of the combinations:

20 * 4 = 80

hi your a little more advanced in the brain than I.


can you explain the factorial process you have for the


combination's for the committee of 3 senior to 1 junior


you have 6!/3! 3!

then for the other you have

only 4!/3!

please explain

Here is the combination formula:

n!/n!(n-r)!

So for the senior officers you have 6 total (n) and you choose 3 (r).

6!/3!(6-3)! = 6!/3!3! = 20

For the junior officers you have 4 total (n) and you choose 1 (r).

4!/1!(4-1)! = 4!/1!3! = 4!/3! = 4

Then you multiply the two for the total number of combination.

20 * 4 = 80


I hope this helps.

Should the formula not be n!/r!(n-r)!

kat187 wrote:hi your a little more advanced in the brain than I.


can you explain the factorial process you have for the


combination's for the committee of 3 senior to 1 junior


you have 6!/3! 3!

then for the other you have

only 4!/3!

please explain

mj41 wrote:Should the formula not be n!/r!(n-r)!

yes, the formula is indeed the one that you have written.

kat187,

dmateer25 has already shown you the correct way, how nCr formula works.

But, here is a shortcut to get this value, without bothering much about factorials.

Here it goes ....

Suppose you have to find nCr ( or you have to SELECT r things out of n) :

for the numerator, write the product of r consecutive integers, beginning with n, each 1 less than the previous.

for the denominator, write the product of r consecutive integers beginning with 1, each 1 more than the previous.

For e.g calculate 6C3 :

step 1 : for the numerator, write the product of 3 consecutive integers, beginning with 6, each 1 less than the previous.

= 6 * 5 * 4 ( don't simplify. Let it be as the way it is).

step 2 : for the denominator, write the product of 3 consecutive integers beginning with 1, each 1 more than the previous.

= 1 * 2 * 3


therefore the value of 6C3 is numerator/denominator = (6 * 5 * 4) / (1 * 2 * 3) = 20

similarly, 4C2 = ( 4 * 3) / (1 * 2) = 6

hope it becomes easier for you now!

[spoiler]hope it becomes easier for you now![/spoiler]

it surely does, man!!