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What else can I help you with?
How many combinations of 12 can you make from 46?
To find the number of combinations of 12 from 46, you can use the combination formula ( C(n, r) = \frac{n!}{r!(n-r)!} ), where ( n ) is the total number of items, and ( r ) is the number of items to choose. For this case, ( C(46, 12) = \frac{46!}{12!(46-12)!} ). Calculating this gives you a total of 16,109,120 combinations.
What is the scope of a general statement of an infinite surd?
I think that you're doing the same thing that I did for IB. Here's what I think you're asking for : 0, 2, 6, 12, 20, 30, 42... c₁ = 0 c₂ = 2 c₃ = 6 (c₂ + 4 = 2 + 4 = 6) c₄ = 12 (c₃ + 6 = 2 + 4 + 6 = 12) c₅ = 20 (c₄ + 8 = 2 + 4 + 6 + 8 = 20) c₆ = 2 + 4 + 6 + 8 + 10 etc... dn = (n/2) <2c₁ + (n-1) 2> dn = (n/2) <2 (2) + (n-1) 2> dn = (n/2) (4 + 2n - 2) dn = (n/2) (2 + 2n) dn = (2n/2) + (2n²/2) dn = n + n²
Does wheel have 2 syllables?
Neither it has 12 You c u n t
How many 4 person can be chosen from a pool of 12 people?
To determine how many groups of 4 can be chosen from a pool of 12 people, you can use the combination formula, which is ( C(n, r) = \frac{n!}{r!(n-r)!} ). Here, ( n = 12 ) and ( r = 4 ). This gives ( C(12, 4) = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} = 495 ). Thus, 495 different groups of 4 people can be formed from the pool of 12.
Given 12 website show many ways can you visit half of them?
To determine how many ways you can visit half of 12 websites, you need to select 6 websites from the total of 12. This can be calculated using the binomial coefficient, which is represented as ( C(12, 6) ). The formula for the binomial coefficient is ( C(n, k) = \frac{n!}{k!(n-k)!} ). Therefore, ( C(12, 6) = \frac{12!}{6!6!} = 924 ). Thus, there are 924 ways to visit half of the 12 websites.
12 n on a c?
12 numbers on a clock(face)
What is '12 N on a C F ' please?
12 Numbers on a Clock Face
How many different ways are there to choose a dozen donuts from the 21 varieties at a donut shop?
The answer depends on if you can choose the same kind of donuts more then once. Or in other words, is repetition permitted. If you can only choose the same kind of donuts only once, it is a 21 choose 12 problem: C(n,k) = n! / (k! (n - k)!) C(21, 12) = 21! / (12! (21 - 12)!) = 21! / (12! (9)!) = 293,930 If you can choose the same kind of donuts more then once, it is a combination with repetition problem. P(n+k-1,k) = (n+k-1)! / (k! (n-1)!) or put it into C(n,k) with n+k-1 as 21 + 12 - 1 = 32 and k as 12 so C(21+12-1,12) = C(32, 12) = 32! / (12! (32 - 12)!) = 32! / (12! (20)!) = 225,792,840
12 N on a C F?
12 Noon on a clock face. 12 Numbers on a clock face.
How many combinations of 12 can you make from 46?
To find the number of combinations of 12 from 46, you can use the combination formula ( C(n, r) = \frac{n!}{r!(n-r)!} ), where ( n ) is the total number of items, and ( r ) is the number of items to choose. For this case, ( C(46, 12) = \frac{46!}{12!(46-12)!} ). Calculating this gives you a total of 16,109,120 combinations.
What elements are there in B-12?
Vitamin B-12 contain Co, C, H, O and N.
What is the scope of a general statement of an infinite surd?
I think that you're doing the same thing that I did for IB. Here's what I think you're asking for : 0, 2, 6, 12, 20, 30, 42... c₁ = 0 c₂ = 2 c₃ = 6 (c₂ + 4 = 2 + 4 = 6) c₄ = 12 (c₃ + 6 = 2 + 4 + 6 = 12) c₅ = 20 (c₄ + 8 = 2 + 4 + 6 + 8 = 20) c₆ = 2 + 4 + 6 + 8 + 10 etc... dn = (n/2) <2c₁ + (n-1) 2> dn = (n/2) <2 (2) + (n-1) 2> dn = (n/2) (4 + 2n - 2) dn = (n/2) (2 + 2n) dn = (2n/2) + (2n²/2) dn = n + n²
Does wheel have 2 syllables?
Neither it has 12 You c u n t
How many 4 person can be chosen from a pool of 12 people?
To determine how many groups of 4 can be chosen from a pool of 12 people, you can use the combination formula, which is ( C(n, r) = \frac{n!}{r!(n-r)!} ). Here, ( n = 12 ) and ( r = 4 ). This gives ( C(12, 4) = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} = 495 ). Thus, 495 different groups of 4 people can be formed from the pool of 12.
What is the nth term of the sequence 6 0 -6 -12?
There are infinitely many polynomials of order 4 (or higher) that will give these as the first four numbers and any one of these could be "the" rule. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.The simplest rule is a linear polynomial U(n) = 6*(2 - n) for n = 1 , 2, 3, ...
Which has more protons in the nucleus C-12 C-14 or N-14?
N-14 has more protons. It contains one more proton than C.
Given 12 website show many ways can you visit half of them?
To determine how many ways you can visit half of 12 websites, you need to select 6 websites from the total of 12. This can be calculated using the binomial coefficient, which is represented as ( C(12, 6) ). The formula for the binomial coefficient is ( C(n, k) = \frac{n!}{k!(n-k)!} ). Therefore, ( C(12, 6) = \frac{12!}{6!6!} = 924 ). Thus, there are 924 ways to visit half of the 12 websites.
