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....

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- Nov 14th 2006, 06:48 AMbriancombo'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.... - Nov 14th 2006, 06:53 AMThePerfectHacker
- Nov 14th 2006, 07:30 AMSoroban
Hello, brian!

Have you had**no**instruction on Permutations and Combinations?

Quote:

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: .$\displaystyle _{20}C_4 \:=\:\binom{20}{4} \:=\:\frac{20!}{4!16!} \:=\:4,845$ 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: .$\displaystyle 20 \times 19 \times 18 \times 18 \:=\:116,280$ possible choices.

But this long list includes choices like: $\displaystyle \{A,B,C,D\}$ and $\displaystyle \{B,D,C,A\}$

. . Since the order is not important, these two selections are identical.

Since 4 objects can be arranged in $\displaystyle 4! = 24$ 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: .$\displaystyle \frac{116,280}{24} \,=\,4,845$