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
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.
36
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.
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.
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.
12 cupcakes
12 hours.
12
12 of them.
how many zeros in 12 lakh