Please help me I really need help?

Question

1) Suppose a contest offers $1000 to the person who can guess the winning four digit number. How many possibilities are there?

2) There are 10 women and 8 men in a club. How many different committees of 6 people can be selected from the group if equal numbers of men and women are to be on the committee?

3)Two marbles are from a bag containing 6 white, 4 red , and 6 green marbles. Find the probability of both marbles being white.

Answer 1

1) 10,000 (there are 10 possibilities for the first digit, 10 possibilities for the second digit, 10 possibilities for the third digit, and 10 possibilities for the fourth digit). 10*10*10*10 = 10,000.

2) 10c3 * 8c3 = [(10*9*8)/(3*2*1)] * [(8*7*6)/(3*2*1)] = 120*56 = 6,720. There are 10 men of which three will be chosen for the committee. There are 10 to pick from for the first man, 9 men remaining to be the second chosen, and 8 remaining to be third. However, choosing Fred then George then Tom is the same as choosing Tom then George then Fred, so you have to divide 10*9*8 by 3*2*1, representing the six combinations of ordering for choosing any three men. That yields 120 different groups of three men are possible. The same calculation on the smaller set of women yields 56 different groups of three women. 120 unique groups of 3 men * 56 unique groups of 3 women = 6,720 possible different committees.

3) (6/16)*(5/15) = 1/8. There are 16 marbles at the start of which 6 are white, so there is a 6/16 (3/8) chance of picking a white marble first. After choosing a white marble first, there are 15 marbles remaining of which 5 are white, so there is a 5/15 (1/3) chance of picking a white marble second after picking a white marble first.

Answer 2

for number 1: Question - I have a combination lock with four places, each has the
number possibility of zero to nine. I want to know how many possible
numbers combinations can fit (example 0000,0001 thru 9999). Also where on
the web would there be a calculator that could answer this for me?
Besides you guys!!
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Since all normal digits are used, there are 10000 possible combinations.
Calculate it by taking the number of possibilities in the first spot, 10,
times the number in the second (10), the third, and fourth: 10*10*10*10 =
10000.

There isn't really any special calculator needed.

Trying all the combinations--just count from 0 to 9999--is fairly easy to do
with such a lock. With only 10000 combinations, it can be done in a few
hours.

Thanks,
Eric Tolman
Computer Scientist
====================================================
Hello,

The first digit can be 0 to 9, i.e., 10 different numbers. The second digit
can also be 9, so you have 10 x 10 =100 combinations (0 to 99) with two
sets of numbers. And so on. With four sets of digits you would have a total
of 10,000 combinations (including 0000).

You would not need a calculator for this one.

Can you think of a possible combination that is not in the 0 to 9999 range?
The answer is no.

Good luck.

Dr. Ali Khounsary
Advanced Photon Source
Argonne National Laboratory
==============================================================
Connie,

If I read your question correctly, you ask how many four digit combination
numbers can be produced using digits 0 through 9. If so, this can be easily
determined by looking at the possibilities:

0000 <--- possibility # 1
0001 through 9999 <--- 9999 more combination possibilities


This gives a total of 10,000 combination of four-digit numbers using the
digits 0 through 9.

A way to calculate this is by N**x
where N is the number of digits possible (here, 10 different digits are
possible) 0 through 9
and x is the number of places available (here we have 4 places available
for a 4 digit number)

note the formula is N raised to the x power

meaning, 10**4

or 10 X 10 X 10 X 10 = 10, 000 combinations


another example: how many combinations are available using digits 0-3 for
a 2 digit number

the possibilities are: 00 01 02 03 10 11 12 13 20 21 22 23 30
31 32 33
or 16 possibilities

the formula would tells us:

N**x is 4**2 or 4 x 4 = 16

Thanks for using NEWTON!


Richard R. Rupnik
Internal Quality Auditor
Lucent Technologies
====================================================================
Actually you just answered you own question by seeing that the numbers went
0000, 0001,....9999.
you can go from 1 to 9999 plus the option for zero. This means you have
9999 + 1 possible combinations or 10,000 possible combinations.
Look in an elementary statistics book to get a better idea about
combinations and permutations. This will give you equations that you can
use on just about any calculator.
BTW, a quick way to figure this is that you have 10 possible numbers (0-9)
to use in 4 different places. Therefore, the number of combinations is 10^4
= 10,000. If you had only 2 numbers to use (0 or 1) in 4 different places,
the number of combinations would be 2^4 = 16. This 2 number system, or
binary system, is the basis for computers and calculators everywhere.

C. Murphy
====================================================================
The easy answer to your question is to think of each possible combination
as a number (which they are). For example, 0000 is zero, 0001 is one, 0020
is twenty, and so on. How many numbers can you make that correspond to
different combinations? 10,000. These are 0001 to 9999, and then 0000 for
the last, so 9999 + 1 = 10000.

This doesn't need a calculator, don't you agree?

Richard Barrans Jr., Ph.D.
Chemical Separations Group
==================================================================

Nice part of the world you live in. Pretty. Green.
There are 10 possibilities for the first number in the combination,
right? For EACH of these numbers, there are 10 possibilities for the
second number, giving a total of 10 x 10 = 100 possible combinations
of 1st and 2nd numbers. Proceeding logically, there are 10^4 = 10,000
combinations for 4 places. Another way to see this quickly is from
your notation above ("0000,0001 . . . 9999"). Pretty clearly, the
available combinations form the numbers between 0 and 9999. How many
such numbers are there? Ten thousand, natch.
If you are thinking of picking the lock, you can see you have a
problem. But all is not lost: cheap combination locks will sometimes
open if you are close but not exactly on the right numbers. That is,
if the first number ought to be 1, the lock will often open if you
dial in 2 or 0, or possibly even 3 or 9. If the lock will open when
you dial in the right number +/- 1, then you actually only have 4
numbers per place (because 1 will work for 0,1,2, 4 will work for
3,4,5, 7 will work for 6,7,8, and then 9 works for 9, gee). There
would then be only 4^4 = 256 combinations to try.
You can also try pulling on the shaft while rotating the wheels (if
it looks like I think it does). Usually one of the wheels will be
holding up the shaft a little more than the others. If you feel very
carefully as you turn the wheels, you may feel the shaft move slightly
out when you hit the right number of that particular wheel.
Continuing in this process, you may be able to pick the lock.

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for number 3:Here are two approaches . . .

There are:(16/2)=16!/2!4!=120 possible pairs of marbles.
There are:(6/2)=6!/2!4!=15 pairs of white marbles.
. . p(both white)=15/120=1/8

The probability that the first marble is white is: 6/16
The probability that the second marble is white is: 5/15
. . p(both white)=6/16 x 5/15=1/8

Answer 3

1)10c4?
2)18c6?
3)16p6?