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.
75 i think
Every number from 100 to 999 inclusive !
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.
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
12! / 9! / 3!
210=1024
The question is exactly equivalent to: "What's the highest binary number with 6 bits ?There are 64, corresponding to the binary numbers from 0 to 63.