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X Roman numeral 10
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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)!]
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.
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).
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
