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“number of bit strings”, so I thought
Let’s think about just placing the 0’s and let the 1’s fill in the remaining spots. The cases we are interested in are how to get exactly 3, 4, 5, 6, and 7 0’s.
How many strings have exactly 3 0’s? This is just the number of ways to choose 3 locations out of 10 available locations where order doesn’t matter, which is $\binom{10}{3}$. Remember, we’re only focusing on the 0’s – the 1’s just fill in the remaining spots.
The argument is the same for getting exactly 4, 5, 6, and 7 0’s. To get the total, we just add them all up:
$$
\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} + \binom{10}{7}.
$$
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combinatorics – A simple "number of bit strings", so I thought

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Sumary: How many bit strings of length 10 contain at least three 1s and at least three 0s?

Matching Result: How many bit strings of length 10 contain at least three 1s and at least three …
 Intro: A simple “number of bit strings”, so I thought Let’s think about just placing the 0’s and let the 1’s fill in the remaining spots. The cases we are interested in are how to get exactly 3, 4, 5, 6, and 7 0’s. How many strings have exactly 3 0’s?…

Source: https://math.stackexchange.com/questions/42400/asimplenumberofbitstringssoithought
Frequently Asked Questions About how many bit strings of length 10 contain at least three 1s and at least three 0s?
If you have questions that need to be answered about the topic how many bit strings of length 10 contain at least three 1s and at least three 0s?, then this section may help you solve it.
How many 10bit bit strings contain more 0s than 1s?
How many bit strings of length 10 have more 0s than 1s? The answer is C(10, 3) = 10. 9. 8 / 3. 2. 1 = b>120/b>.
How many different 10bit bit strings are there?
In total, there are 386 bit strings of length 10 that contain more 0s than 1s (210 + 120 + 45 + 10 + 1 = 386). c) To determine how many contain seven or more 1s, we must count the strings that contain seven, eight, nine, or ten of them.
How many 10bit bit strings start with 3 zeros or end with 3 ones?
Total: (25) x (21) x (22) = 32 x (22) x (44) = 256.
How many 10bit strings have beginning and ending 1s?
256bit strings
How many 10 bit strings have a maximum of three 1s in them?
There are at least three 1s and at least three 0s in each of the 912 strings of length 10.
How many 10 character strings have exactly four 1s in them?
This simply asks us to select 4 out of 10 slots to place 1s in, so the answer is: = 9!/4! = 15,120. a) exactly four 1.
How many bit strings of length 10 have a 00 or 111 at the beginning or end?
Strings of length 10 either begin with 000 or end with 111: 1*1*1*1*2*2*2*2*2*2*2*2=128; 128+12816=240.
How many 15bit bit strings have a minimum of three 1s and three 0s?
The number of strings is 215 – 1 – 1 – 15 – 15 – 105 – 105, which equals 32526.
How many 12 bit strings have a maximum of three 1s in them?
three 1s maximum?) = 1 + 12 + 66 + 220 = 299
How many 10digit ternary 0/1 or 2 strings exist that contain precisely two 0s, three 1s, and five 2s?
17) Given a string of 10 ternary digits, where each digit is either a 0 or a 1, and where there are exactly two 0s, exactly three 1s, and exactly five 2s, how many such strings exist? The answer is 10!
How many binary strings with 10 digits that start or end in 1 can you name?
The inclusionexclusion formula yields A?B=A+B?A?B=29+29?28=768 for the number of 10digit binary strings that start or end in 1.
How many 10digit 0 and 1number sequences exist that don’t have any backtoback 0s?
The answer is 144 for 10digit sequences of 0s and 1s devoid of any backtoback 0s.