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.
56 The number of triples of 1s on 8 bits
There are 210.
Every number from 100 to 999 inclusive !
75 i think
length/width
56 The number of triples of 1s on 8 bits
Three of them ... the binary numbers 3, 5, and 6. 011 101 110
There can be only one.
There are 27 of them.
Only 1 exists, and it is "999"
One: 444!
-- 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.
12! / 9! / 3!
Every bit can either be a 0 or 1. So to find the amount of bit strings of length either, you do 2length to find the amount of bit strings there are of a given length.
1024
To find the number of strings of length 5 generated by the productions S → a0b, a → bb0, and b → aa1, we can analyze the derivations. Starting with S, we can replace it with a0b or explore the substitutions for 'a' and 'b'. Each substitution increases the length of the string, so we need to systematically track how the productions can be combined to form strings of exactly 5 characters. After evaluating the combinations, we find that the total number of strings of length 5 is 8.
There are (2^5) bit strings of length 5, as each bit can be either 0 or 1. Therefore, the total number of bit strings is (32).