Find the point on the parabola y^2 = 8x at which the ordinate is double the abscissa.
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The number of possibilities for constructing the 4 characters long password with specified conditions is given by: Option C: 703,040
<h3>What is the rule of product in combinatorics?</h3>If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in ways.
Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.
Thus, doing A then B is considered same as doing B then A
We're specified that:
- The password needs to be 4 characters long
- It must have 3 letters and 1 of 10 special characters.
- Repetition is allowed.
So, each of 3 characters get 26 ways of being 1 letter. (assuming no difference is there between upper case letter and lower case letter).
And that 1 remaining character get 10 ways of being a special character.
So, by product rule, this choice (without ordering) can be done in:
ways.
Now, the password may look like one of those:
- L, L, L, S
- L, L, S, L
- L, S, L, L
- S, L, L, L
where S shows presence of special character and L shows presence of letter.
Those 175760 ways are available for each of those four ways.
Thus, resultant number of ways this can be done is:
Thus, the number of possibilities for constructing the 4 characters long password with specified conditions is given by: Option C: 703,040
Learn more about rule of product here: