There are 10C3 = 10*9*8/(3*2*1) = 120 combinations.
It is 10C3*(1/2)10 = 10*9*8/(3*2*1)*(1/1024) = 0.1172, approx.
its solve easy
to solve them you have to do stagigys
solve it
The answer depends on the value of c!
There are: 10C3 = 120
There are 10C3 = 10*9*8/(3*2*1) = 120 combinations.
To calculate the number of 3-digit combinations that can be made from the numbers 1-9, we can use the formula for permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the total number of options (10 in this case) and r is the number of digits in each combination (3 in this case). Therefore, the total number of 3-digit combinations that can be made from the numbers 1-9 is 10^3 = 1000.
10C3(0.5)^10
It is 10C3*(1/2)10 = 10*9*8/(3*2*1)*(1/1024) = 0.1172, approx.
solve solve solve
just solve it yes you have to solve but he is asking you how to solve and also what are the steps to solve the specific answer.
you solve it
its solve easy
to solve them you have to do stagigys
Generally,1. Convert parallel branches into series equivalents2. Solve for the total resistance3. Solve for individual voltages4. Solve for individual currents5. Solve for power