Q: How many combinations r there in 4 colors?

Write your answer...

Submit

Still have questions?

Continue Learning about Other Math

The number of combinations of r numbers out of n is nCr = n!/[r!*(n-r)!] where n! = 1*2*3...*n So here n = 30, r = 5 which gives 30*29*28*27*26/(5*4*3*2*1) = 142,506

nCr=n!/r!/(n-r)!

First, let us consider each case separately.Looking at the first case, we need to find how many ways we can choose 5 sopranos from 7 sopranos. When we need to find the amount of combinations of n objects when we can only choose r objects, we can use the formula:combinations = n!/r!(n-r)!The ! symbol means to multiply that number by all the numbers before it.So 4! = 4 * 3 * 2 * 1 = 24n = the total number of sopranos we can choose from (7 sopranos)r = the amount of sopranos we can actually choose (5 sopranos)So plugging everything into the formula:combinations = 7!/5!(7-5)! = 7!/5!*2! = (7*6*5*4*3*2)/(5*4*3*2)*(2) = 21 different combinations.Next, let's consider the altos.Our formula (combinations = n!/r!(n-r)!) is the same so we just need to plug in a different set of values.n = the total number of altos we can choose from (9 altos)r = the amount of altos we can actually choose (4 altos)So, using the formula:combinations = 9!/4!(9-4)! = 9!/4!*5! = (9*8*7*6*5*4*3*2)/(5*4*3*2)*(4*3*2) = 126 different combinations.If we treat the choosing of the alto's and soprano's as two independent events, then we can figure out the number of combinations of both events by multiplying the number of combinations for both. In this case, we would multiply the number of combinations for choosing 5 from 7 sopranos (21) by the number of combinations for choosing 4 from 9 altos (126.) This gives us 21 * 126 = 2646 different combinations.

If you have N things and want to find the number of combinations of R things at a time then the formula is [(Factorial N)] / [(Factorial R) x (Factorial {N-R})]

Oh, what a happy little question! To find the number of combinations in 21 numbers, we can use a formula called combinations. It's like mixing colors on your palette - each combination is unique and beautiful in its own way. So, for 21 numbers, the number of combinations would be calculated using the formula nCr = n! / r!(n-r)!. Just remember, there's no mistakes, just happy little accidents in math!

Related questions

10,000 * * * * * WRONG! That is the number of permutations, NOT the number of combinations. The number of combinations denoted by nCr = n!/[r!*(n-r)!] = 10!/[4!*6!] = 10*9*8*7/(4*3*2*1) = 210

The number of R-combinations in a set of N objects is C= N!/R!(N-R)! or the factorial of N divided by the factorial of R and the Factorial of N minus R. For example, the number of 3 combinations from a set of 4 objects is 4!/3!(4-3)! = 24/6x1= 4.

10,000

There are 120 combinations of 5 items, which can be calculated using the formula nCr = n! / r!(n - r)!.

Just use combinations formula. nCr, where n=44, r=6. Plug it into the calculator or use the formula, nCr = n!/[r!(n-r)!] And you should get 7059052 as the number of combinations.

The number of combinations of r numbers out of n is nCr = n!/[r!*(n-r)!] where n! = 1*2*3...*n So here n = 30, r = 5 which gives 30*29*28*27*26/(5*4*3*2*1) = 142,506

If you have n objects and you are choosing r of them, then there are nCr combinations. This is equal to n!/( r! * (n-r)! ).

Formula: nCr = n! divided by (n-r)! x r!where n is the number of things to choose from and you choose r of them15C4 = 15!divided by (15-4)! x 4! = 1365 ways

nCr = n!/((n-r)!r!) → 49C8 = 49!/((49-8)!8!) = 49!/(41!8!) = 450,978,066 combinations.

To determine the number of combinations possible using the numbers 9, 3, 1, and 7, we can use the concept of permutations. A permutation is an arrangement of objects in a specific order. In this case, we want to find the number of permutations of all the numbers. The formula for permutations of n objects taken r at a time is given by: P(n, r) = n! / (n - r)! Where "!" denotes the factorial function. In this case, we have 4 numbers and we want to arrange all of them, so r = 4. P(4, 4) = 4! / (4 - 4)! = 4! / 0! = 4! / 1 = 4 * 3 * 2 * 1 = 24 Hence, there are 24 different number combinations that can be made using the numbers 9, 3, 1, and 7.

nCr=n!/r!/(n-r)!

Try 151,600! Permutations & Combinations. P(n,r)=n!(nār)! not P(n,r)=n!/(n!-r!)r! ?