How many samples of size n=3 can be drawn from this population?.Let N be the total number of times until a number larger than 4 is observed. I roll a fair die repeatedly until a number larger than 4 is observed.What is the z-value that will give an area of 0.975 on the right side of the normal curve? The amount of coffee dispensed in a dispensing machine is normally distributed with a mean of 190 ml and a standard deviation of 10 ml.Therefore the total number of permutations is 720\times 42=30240 720 × 42 = 30240 permutations. Now, the number of permutations for the second to the seventh letter (between the first and the last letter) is 6\times5\times4\times3\times2\times1=720 6 × 5 × 4 × 3 × 2 × 1 = 720 permutations. In the second position, we can choose from any of the 6 letters thus, there are 6 ways to choose the six letters to occupy the second positionįor the third position, there are 8-3=5 8 − 3 = 5 letters to choose from to occupy this position thus there are 5 ways to choose the 5 letters to occupy the third position.įor the fourth position, there are 8-4=4 8 − 4 = 4 letters to choose from to occupy this position thus there are 4 ways to choose the 4 letters to occupy the fourth position.įor the fifth position, there are 8-5=3 8 − 5 = 3 letters to choose from to occupy this position thus there are 3 ways to choose the 3 letters to occupy the fifth position.įor the sixth position, there are 8-6=2 8 − 6 = 2 letters to choose from to occupy this position thus there are 2 ways to choose the 2 letters to occupy the sixth position.įor the seventh position, there is 8-7=1 8 − 7 = 1 letter remaining to occupy this position thus there is only 1 way for the remaining letter to occupy this position. The number of permutations for the first and last letters is 7\times 6=42 7 × 6 = 42 permutations.Įxcluding the letters in the first and last positions, we remain with 8-2=6 8 − 2 = 6 letters. Thus the number of choices is 8-2=6 8 − 2 = 6 The reason is that for this position, we exclude the letter H and the letter occupying the first position. The number of choices available for the last letter is 6. Thus, we got 7 choices for this position. The reason is that, out of the 8 letters only letter A cannot occupy this position. The number of choices available for the first letter is 7. We want an arrangement which does not begin with A and does not end with H.įirst, let us determine the number of possible choices for the first and last letters.
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