Basically, this is the same as finding the number of distinct ways of arranging seven 1s and 3 0s. That is (10!/7!3!) = (10*9*8)/(3*2*1) = 120. There are 120 bit strings of length 10 with exactly three 0s.
It looks exactly like an Isosceles triangle, because two of its three sides have the same length.
The length of the longer leg of a right triangle is 3ftmore than three times the length of the shorter leg. The length of the hypotenuse is 4ftmore than three times the length of the shorter leg. Find the side lengths of the triangle.
A trinomial
it is...........two divided by three
Basically, this is the same as finding the number of distinct ways of arranging seven 1s and 3 0s. That is (10!/7!3!) = (10*9*8)/(3*2*1) = 120. There are 120 bit strings of length 10 with exactly three 0s.
Three of them ... the binary numbers 3, 5, and 6. 011 101 110
There can be only one.
How many strings of three digits are there? 000 to 999, or a total of 1000. How many strings of three digits contain the same three digits? That's 000, 111, 222 ... 999! ten in total. The difference is your answer: 1000-10 = 990.
There are 27 of them.
Only 1 exists, and it is "999"
One: 444!
Five (5) have one or the other but not both. Six (6) have both. Total of eleven (11).
-- 893 of them have no 4's. -- Three of them have exactly one 4: 4xx, x4x, and xx4. -- Three of them have exactly 2 4's: 44x, 4x4, and x44. -- One of them has exactly three 4's: 444. -- None has more than three 4's.
How do you do a boondoggle with three strings
Exactly one.
A viola does not have three strings in fact it has four