do you mean how many combinations of any size? So we have 12 combinations of size 1 12C2=12x11/2 =66 of size 2 12C3 of size 3 and we keep going and need to add these. But there is a formula for adding all these. Think of the each number between 1 and 12, say 1, for any combination it is either in that combination of not in it. So we have 2^12 combinations. That is 4096
12, including a combination containing 0 digits.
Assuming no repetition, 12 x 11 x 10.... x 4 x 3 x 2. If repetition is allowed, 1212
To determine how many different triangles can be formed with 12 points on a circle, we can use the combination formula. Specifically, we need to choose 3 points from the 12, which is calculated as ( \binom{12}{3} ). This gives us ( \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 ). Therefore, 220 different triangles can be formed with 12 points on a circle.
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.
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The 12V battery connected to the 2.4 Ohm combination will supply 12/2.4 or 5A. The individual currents will be 12/3 or 4A for the 3 Ohm resistor, and 12/12 or 1A for the 12 Ohm resistor. The 2.4 Ohm parallel combination is obtained from a simple product-over-the-sum calculation.
The number of combinations of 12 numbers taken 12 at a time (i.e., choosing all 12 numbers from a set of 12) is calculated using the binomial coefficient formula, which is ( \binom{n}{k} = \frac{n!}{k!(n-k)!} ). For ( n = 12 ) and ( k = 12 ), this simplifies to ( \binom{12}{12} = 1 ). Therefore, there is only one combination of 12 numbers from 1 to 12, which includes all the numbers themselves.
To determine how many ways an adviser can choose 4 students from a class of 12, we use the combination formula, which is given by ( C(n, r) = \frac{n!}{r!(n-r)!} ). Here, ( n = 12 ) and ( r = 4 ). Thus, the calculation is ( C(12, 4) = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} = 495 ). Therefore, there are 495 ways for the adviser to choose 4 students from the class.
A 12-lead ECG uses a total of 10 electrodes. These include six chest (precordial) electrodes and four limb electrodes. The combination of these electrodes allows for the recording of 12 different electrical views (leads) of the heart's activity.
12 cupcakes
Combination 1 = 0 Pigs 15 Hamsters 5 Rabbits Combination 2 = 4 Pigs 12 Hamsters 4 Rabbits Combination 3 = 8 Pigs 9 Hamsters 3 Rabbits Combination 4 = 12 Pigs 6 Hamsters 2 Rabbits Combination 5 = 16 Pigs 3 Hamsters 1 Rabbits
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