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What does c mean in math Example 5c3 and 3c2?
it means combinatorial or combination, you use the formula nCr = n!/((n-r)! x r!). example 5c3 = 5! / ((5-3)! x 3!) = 5! / (2! x 3!) = (5x4x3x2x1) / ((2x1) x (3x2x1)) = 10
How do you find the coefficients of the terms in the binomial expansion?
The coefficient of x^r in the binomial expansion of (ax + b)^n isnCr * a^r * b^(n-r)where nCr = n!/[r!*(n-r)!]
What is geometric terms?
Geometric Sequences work like this. You start out with some variable x. Your multiplication distance between terms is r. Your second term would come out to x*r, your third x*r*r, and so on. If there are n terms in the sequence, your final term will be x*r^(n-1).
Ten times the square of a non-zero number is equal to eighty times the number What is the number?
10 x n x n = 80 x n Divide by 10 x n; n = 8 Job done.
How can you figure out combinations in math?
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})]
How many different 3 digit combinations can there be if the digits are distinct?
With 3 digit numbers there are 10 digits (0, 1, ..., 9) from which to select 3 different digits. This can be done in: 10C3 = 10! / (10-3)!3! = 120 ways. For a set with n items, r of them can be selected (permuted) in: n x (n-1) x ... x (n - (r +1)) = n! / (n-r)! = nPr ways. However, if the order of selection does not matter (just the combination of the r items) then the selected items can be chosen in r x (r-1) x ... x 1 = r! ways. Meaning there are nPr / r! = nCr = n! / (n-r)!r! combinations possible.
How many combinations of 8 numbers are possible from 10 numbers?
In general, the number of combinations of n things taken r at a time isnCr = n!/[(n - r)!r!]Thus, we have:10C8= 10!/[(10 - 8)!8!]= 10!/(2!!8!)= (10 x 9 x 8!)/(8! x 2 x 1)= (10 x 9)/2= 5 x 9= 45
How many ways can a committee of 3 people be chosen out of 10 people?
Combination of three people from a group of 10 = (n!)/[r! x (n-r)!], where n = 10 and r = 3. The answer is therefore (10.9.8.7.6.5.4.3.2.1)/[(1.2.3) x (1.2.3.4.5.6.7)] = (10.9.8)/(1.2.3) = 10.3.4 = 120 ways
How many combinations of 3 out of ten numbers?
There are 10 x 9 x 8 ways to select 3 numbers from 10 in order (permutations of 3 numbers from 10), but as want the number of combinations, the order doesn't matter, so need to divide by the number of ways 3 numbers can be ordered which is 3 x 2 x 1. Thus: combinations(3 from 10) = (10 x 9 x 8)/(3 x 2 x 1) = 120. There are formulae for the number of permutations and combinations of r items chosen from a set of n items: Permutations: nPr = n!/(n-r)! Combinations: nCr = n!/(n-r)!r! Where n! means "n factorial" which is the product of n with all numbers less than it to 1, ie n! = n x (n-1) x ... x 2 x 1 The combinations nCr is also written as a vector with n over r, as in (nr), but it is difficult to show the n exactly over the r as it should be (with the characters available) so I use the nCr format which is found on scientific calculators.
Formula for EMI?
E= P x r x (1 + r)n / ((1+r)n -1) Here p=principal amount r = interesr rate per month(ex: if interest rate per annum is 10% then 10/(12*100)) n= tenure in months
What is 10 x n 0.06 then what is n?
10 x n = 0.06 /10 /10 n = 0.006
What is the general term of the expansion 1-x to the power n?
The r+1 th term is nCr(-x)r where r = 0, 1, 2, ... , n. and where nCr = n!/[r!*(n-r)!]
How do you find all the combinations of a list?
To find the number of combinations possible for a set of objects, we need to use factorials (a shorthand way of writing n x n-1 x n-2 x ... x 1 e.g. 4! = 4 x 3 x 2 x 1). If you have a set of objects and you want to know how many different ways they can be lined up, simply find n!, the factorial of n where n is the number of objects. If there is a limit to the number of objects used, then find n!/(n-a)!, where n is the number of objects and n-a is n minus the number of objects you can use. For example, we have 10 objects but can only use 4 of them; in formula this looks like 10!/(10-4)! = 10!/6!. 10! is 10 x 9 x 8 x ... x 1 and 6! is 6 x 5 x ... x 1. This means that if we were to write out the factorials in full we would see that the 6! is cancelled out by part of the 10!, leaving just 10 x 9 x 8 x 7, which equals 5040 i.e. the number of combinations possible using only 4 objects from a set of 10.
What is the answer in 2 x 5 x 10 - 1?
2 x 5 x 10 - 1= n 10 x 10 - 1=n 100 - 1 =n 99= n
The 10 letter word that has the letters r-a-o-a-e-x-n-t-i-l?
relaxation
What is the binomial probability formula?
T r+1 = (n / r) (a ^n-r) x (b)^r
What does c mean in math Example 5c3 and 3c2?
it means combinatorial or combination, you use the formula nCr = n!/((n-r)! x r!). example 5c3 = 5! / ((5-3)! x 3!) = 5! / (2! x 3!) = (5x4x3x2x1) / ((2x1) x (3x2x1)) = 10
