1. combo's

how many diffrent combonations of 4 can I make with 20 numbers
EX:1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18-19-20

1-2-3-4 AND HOW WOULD I put them into combos
1-2-3-6
ETC....

2. Originally Posted by brian
how many diffrent combonations of 4 can I make with 20 numbers
EX:1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18-19-20

1-2-3-4 AND HOW WOULD I put them into combos
1-2-3-6
ETC....
It depends whether,

1,2,3,4 is same as 1,2,4,3 or different.

If different then the answer is $N=20\cdot 19\cdot 18\cdot 17$. Otherwise,
$N/24$

3. Hello, brian!

Have you had no instruction on Permutations and Combinations?

How many different 4-number combinations can be made from a set of 20 numbers?

Ex: 1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18-19-20

1-2-3-4, 1-2-3-6, etc.

Identical problem: There are 20 different CDs on sale.
. . . . . . . . . . . . . You have enough money to buy four of them.
. . . . . . . . . . . . . In how many ways can you select four CDs?

Since you are choosing the CDs and tossing them in a shopping bag,
. . the order of the CDs is not important.
Hence, this is a "combinations" problem.

The Combination Formula says: . $_{20}C_4 \:=\:\binom{20}{4} \:=\:\frac{20!}{4!16!} \:=\:4,845$ ways.

The derivation goes like this:

You have 20 options for your first choice,
. . . .and 19 options for your second choice,
. . . .and 18 options for your third choice,
. . . .and 17 options for your fourth choice.

It seems that you have: . $20 \times 19 \times 18 \times 18 \:=\:116,280$ possible choices.

But this long list includes choices like: $\{A,B,C,D\}$ and $\{B,D,C,A\}$
. . Since the order is not important, these two selections are identical.

Since 4 objects can be arranged in $4! = 24$ different orders,
. . our number is too large by a factor of 24.
. . Our list has 24 times as many choices as it should have.

Hence, the corrected answer is: . $\frac{116,280}{24} \,=\,4,845$