0
Add your answer:
Earn +20 pts
How many ways can 3 books be selected from 5 books?
(5x4x3)/(3x2x1) = 10
What is the LCM of 4 5 6 10?
It is not possible to give a sensible answer to this question. The least common multiple (LCM) refers to a multiple that is COMMON to two or more numbers. You have only one number in the question!
How many different ways can 5 bicycles be parked if there are 3 available parking spaces?
The permutations and combinations can be calculate in this way:The first spot can be filled 5 ways.The second spot can be filled 4 waysThe third spot can be filled 3 ways.There are a total of 5x4x3=60 ways.
How do you calculate reverse forecast trebles?
It is the same as trebles. Start with how many selections there are and then times by the next number in the sequence and times that by the next number. Then divide that number by 1x2x3. For example 5 selections 5x4x3 =120 / 1x2x3 =6 = 20 trebles in 5 selections. But because it is a reverse forcadt you need to then times that by 2
How do you get probability of 4 digit number is greater than 3251 but less than 8825 using only 1 2 3 5 7 9?
The first thing to do is find how many possible combinations of those numbers match the criteria. All numbers that start with 7 match it. And there are 5x4x3=60 possible numbers that start with 7. All numbers that start with 5 also match it, and there are 5x4x3=60 numbers that start with 5. All numbers that start with 35, 37 or 39 match it. And there are 4x3x3=36 numbers that do this. And finally, all numbers that start with 325, 327 and 329 match it, and there are 3x3=9 numbers that do this. In total this is 60+60+36+9=165 possible 4 digit numbers that match the criteria. The number of possible four digit numbers in total is 6x5x4x3 = 360. This means the probability of a 4 digit number using only 1,2,3,5,7 and 9 being greater than 3251 but less than 8825 is 165/360 which can be simplified to 11/24.
How many 3 digit numbers can be formed with digits 1 2 3 4 5 if no digit occurs twice within the same number?
60. If any of the numbers 1-5 can be in the first position, there are 5 possibilities for that spot. Since whichever number is in the first position can't be used again, that leaves 4 numbers possible for the second spot. Again, since whatever is in the middle can't be used for the last number, that leaves 3 possibilities. 5X4X3= 60
How many three digit numbers can you make using the digits using 5 different numbers if no digit appears more than once in a number?
I'm going to restate the question so you know for sure what question I'm answering, because I'm not positive I know what question you're trying to ask. Given five distinct objects, how many ways is it possible to select three of them if order matters? There are 5 choices for the first digit/object. There are 4 choices for the second digit/object. There are 3 choices for the third digit/object. So there are 5x4x3 = 60 possible combinations.
How many ways can you put 5 different letters in order?
Let's start with the first. You have 5 to choose from so that's 5. Now for each of those 5, let's choose the second letter. You now have 4 to choose from. So we're up to 5 x 4. In the third position we're down to a choice of 3, so total possibilities go up to 5x4x3. In the fourth, we have 2 left, so the choices go up to 5x4x3x2. After all that, there is only one left for the fifth position: just one to choose from. So the final answer: 5x4x3x2x1. That's called 5 factorial.
A test has 5 multiple choice questions each with 4 choices what is the probability of guessing exactly 3 out of 5 right?
Since there are 4 choices the probability of guessing any given answer correctly is 1/4 or .25; call this a success and denote it by p The chance of guessing wrong is .75; call this a failure and denote it by q. So the chance of 3 out of 5 correct answers is 5C3xp^3q^(5-3)=10p^3q^2 5C3x(.25)^3(.75)^2 5x4x3/3x2(.15625)(.5625) 10(.12625)(.5625)=.0877891
How many different way can 5 people sit at a table?
This question is entirely mathematical and i will answer it the best I can in terms that can be easily understood.This scenario can be solved easily using a permutation, a permutation describes the arrangement of objects (or events) in a specific order.For example: ABC is different to ACB, CAB, BAC, BCA and CBA.Generally the number of ways of arranging r objects from a group of n objects is:nPr Which equals n! divided by (n-r)!Note: ! =factorial (4!=4x3x2x1, 8!=8x7x6x5x4x3x2x1) and so onHowever in a circle we have a continuous loop where objects can be arranged. We must take into consideration the direction when dealing with circles.If we differ between clockwise and anticlockwise then the reult would be (n-1)!In this case it would (5-1)!=4!=24However if no distinction is made between clockwise and anticlockwise then the result is (n-1)! divided by 2Therefore the answer would be 4!/2=24/2=12So 5 people can site around a table either 12 different ways, or 24 different ways depending on whether the direction is taken into consideration.
How do you calculate the payoff on a Trifecta at a pari-mutuel Horse Track?
Generally: Take the total amount wagered in the trifecta pool and multiply it by the takeout rate (a.k.a. commission rate). Subtract this amount from the total amount wagered in the pool. From the amount remaining (the Net pool), subtract the amount of money that was placed on the winning Trifecta combination, leaving you with the Profit pool. Divide the Profit pool by the amount that was bet on the winning combination. Add 1 to the division's resulting amount. Drop the pennies so you're left with the dollars and dimes. This is the payout (price) amount on the winning combination per $1 of winning money.
