18 different combinations. When a coin is tossed twice there are four possible outcomes, (H,H), (H,T), (T,H) and (T,T) considering the order in which they appear (first or second). But if we are talking of combinations of the two individual events, then the order in which they come out is not considered. So for this case the number of combinations is three: (H,H), (H,T) and (T,T). For the case of tossing a die once there are six possible events. The number of different combinations when tossing a coin twice and a die once is: 3x6 = 18 different combinations.
We select 1 shirt out of 6 shirts in (6 choose 1) ways or 6 ways. Then, we select 1 out of 3 pairs of shorts in (3 choose 1) or 3 ways. Therefore, the possible combinations of a shirt and a pair of shorts is 6 * 3 = 18 possible combinations.
The chance is 1/36. (There are 36 possible combinations for two 6-sided dice, but only 18 separate combinations when the dice are not considered seperately.)
With a single die:There are 6 sides on a dice.There are 3 sides that are odd (1, 3, 5)3/6 = 1/2 = 50%Therefore, a single die has a 50% chance of landing on an odd number.If you have two dice:I'll explain this the longer way, however there are short (and simpler explanations) that basically use the fact that a single die is 50%, therefore two independent dice will also be 50%.Number of dice = 2Sides of a dice = 6Odd sides: 1, 3, 5, so Number of odd sides on a dice = 3Even sides: 2, 4, 6, so Number of even sides = 3Total combinations with two 6 sided dice = (sides on dice x sides on dice) = 6 x 6 = 36We need to take the fact that:An odd number + an odd number = evenAn even number + an even number = evenAn odd number + an even number = odd
It is possible to choose 4 radios out of 23 in (23 choose 4)=8855 ways. The number of those combinations where there are exactly 2 indefective and 2 defective is: (18 (the number of those indefective) choose 2)*(5 (the number of defective radios) choose 2)=1530 divided by 8855=1530/8855=306/1771 which is 17.2784%.
There are 18C4 = 18!/[18-4)!4!] = 18*17*16*15/(4*3*2*1) = 3060 combinations.
The number of combinations is 20C5 = 20!/(15!*5!) = 20*19*18*17*16/(5*4*3*2*1) = 15,504
233 = 18 combinations.
918 divided by 18 equal 51.
There are 18*17*16/6 = 816 of them!
18 different combinations. When a coin is tossed twice there are four possible outcomes, (H,H), (H,T), (T,H) and (T,T) considering the order in which they appear (first or second). But if we are talking of combinations of the two individual events, then the order in which they come out is not considered. So for this case the number of combinations is three: (H,H), (H,T) and (T,T). For the case of tossing a die once there are six possible events. The number of different combinations when tossing a coin twice and a die once is: 3x6 = 18 different combinations.
Three possible combinations: 17+1, 13+5 and 7+11.
Argon has 18 electrons because it possesses 18 protons in its nucleus, making it electrically neutral. The number of electrons in an atom is equal to the number of protons in the nucleus.
18, since 18/18 = 1
It is: 414/18 = 23
18
18