FAQ

How Many Bit Strings Of Length 10 Contain At Least Three 1s And At Least Three 0s?

912 strings

What are bit strings?

A bit-string is a series of binary digits (bits). The variety of bits withinside the series is known as the period of the value. A bit-string of 0 periods is known as a null string. … A bit string occupies a complete byte.

How many bit strings of length 10 have an equal number of 0s and 1s?

(d) An equal number of 0s and 1s in a string of length 10, means that there are five 0s and five 1s.

How many bit strings of length 10 have at least four ones?

c) at least four 1s? We subtract from the total number of bit strings of length 10 those that have only 0, 1, 2 or 3 1s. 210 − [C(10,0) + C(10,1) + C(10,2) + C(10,3)] = 1024 − 1 − 10 − 45 − 120 = 848.

How many 10 digit bit strings are there?

The first digit can be any of 5 digits, but the other two can be any of 10 digits, so the answer is 5 · 10 · 10 = 500.

How many bit strings of length 12 contain an equal number of 0s and 1s?

How many bit strings of length 12 contain? An equal number of 0s and 1s in a string of length 12, means that there are six 0s and six 1s.

How many 10 bit strings contain 6 or more 1’s?

386 strings

6. How many 10-bit strings contain 6 or more 1’s? Answer: (106)+(107)+(108)+(109)+(1010)=386 strings.

How many bit strings contain exactly five 0s?

How many bit strings contain exactly five 0s and fourteen 1s if every 0 must be immediately followed by two 1s? What I need help with: For this question, the answer is 126-bit strings.

How many 19 bit strings contain at least 4 zeros?

(C) is similar to (b), but “at least four” means four or more. Thus you want to calculate the number of bit strings that have four, five, six,…, up to 10 ones, then add these together (or is there an easier way to do it if we realize that at most three is the opposite of at least four?)

How many bit strings are there of length Ten both begin and end with a 1?

256 bit strings
256 bit strings of length ten starting and ending with a 1.

How many binary strings of length n which is not starting from 10 are possible?

So if you sum all of that up, you get 512 strings which makes sense because 210=1024..

How many bit strings of length 8 or less are there?

There are 511 bit strings of length 8 or less.

How many bit strings of length n contain exactly r 1s?

A bit string of length n with exactly r 1’s will have exactly n−r 0’s.

How many different bit strings of length n are there?

Since a bit consists of either the number 1 or 0, there are only two ways that the first slot can be filled or 2n ways. Since there are eight bits, the answer would be 28 or 256.

How many strings of length 10 over the alphabet ABCD have exactly 3 A’s?

Thus the number of strings of length 10 that have exactly 3 a’s is C(10,3)×27.

How many Bitstrings that is strings of 0s and 1s of length 10 are there that contain an equal number of 0s and 1s?

We add up the number of bit strings of length 10 that contain zero 1s, one 1, two 1s, three 1s, and four 1s. 1 + 10 + 45 + 120 + 210 = 386.

How many bit strings of length 14 have equal number of zeros and ones?

Answer is 2184. (b) Since it’s a bit string of 14 digits and there can be only digits 0 or 1, we need to select 7 0s and the rest would be 7 1s, so the number of permutations would be equal to.

What is an 8 bit string?

A byte is a string of 8 bits.

How many 8 bit strings have weight 5 ie contain exactly five 1s and start with the sub string 101?

There are (53)=10 eight-bit strings of weight five starting with 101 (your A)

How many bit strings of length 8 either start with 01 or end with 01?

4 Answers. We interpret starts with 1 or ends in 01 as meaning that bit strings that satisfy both conditions qualify. By your correct analysis, there are 27 bit strings that start with 1.

How many bit strings are there of length 14 that have exactly five zeros?

How many bit strings contain exactly five 0s and 14 1s if every 0 must be immediately followed by two 1s? Since given bit string contains five 0s and fourteen1s, so the required bit string length is 19.

How many strings can be formed by ordering the letters salespersons if no two s’s are consecutive?

possible strings of SALESPERSONS with no consecutive S’s. Thus there are 2,540,160 strings of SALESPERSONS with no consecutive S’s..

How many binary strings of length 10 are there with at least three 1s?

912 strings

912 strings of length 10 contain at least three 1s and at least three 0s.

How many 8 bit binary strings have exactly 3 zeros?

How many 8-bit strings contains three 0’s in a row and five 1’s? Answer = 6!/5!

How many strings are there of lowercase letters of length 4 or less?

475,254 strings (excluding the empty string)

How many answer patterns are there in a standard multiple choice exam with 10 items?

Since we have two sequential tasks, we use the product rule. Therefore, there are 18 × 325 = 5,850 ways. A multiple-choice test contains 10 questions. There are four possible answers for each question.

How many different three-letter initials contain at least one A?

676 different three-letter initials beginning with an A.

How many different 3 letter initials are there that begin with an A?

676 different three

2 Answers. 676 different three-letter initials are there that begin with an A.

How many length 10 strings are there of the letters a B and C that contain exactly 4 as?

Thus, our final answer is 1+8+40+160=209 possible strings.

How many bit strings of length 9 are there?

How many bit strings of length 9 contain exactly three 1s? 10∗10∗10∗96=531441000 But then those first 1′s don’t necessarily have to be the first 3 digits. They can be elsewhere in the digit string as well.

How many strings of length 6 over a/b/c d are there that have at least one D?

How many strings of length 6 over the alphabet {A, B, C, D, E} have at least 4 consecutive A’s?

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How many different bit strings of length seven are there end with 10?

I am thinking for the strings that start with 10, we would have 7−2=5 bits to choose, so 32 possible bit strings of length 7 that starts with 10.

How many different bit strings of length 3 are there?

The only string of length 3 possible is “111”. The 3 strings are “1110”, “0111” and “1111”.

How many bit strings of length 8 contain either 3 consecutive zeros or 4 consecutive 1s?

Number of binary strings of length 8 that contain either three consecutive 0s or four consecutive 1s. How many bit strings (Consists of only 0 or 1) of length 8 contain either three consecutive 0s or four consecutive 1s? I am getting answer 256 but the provided answer is 147 .

How many bit strings of length 8 either start with a 1 or end with a 00?

Hence, the number of bit strings of length 8 that will either start with 1 or end with 00 is 160.

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

Permutation and Combination:

A permutation refers to rearranging a hard and fast of numbers in a positive way. A mixture may be very similar; however, it refers to continually arranging this set in an ordered form.

912 strings contain at least three ones and at least three zeros.

The total length is = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 =1024.

String with no ones= 10!0!(10−0!)=1

String with no zeros = 10!0!(10−0!)=1.

String with one one = 10!1!(10−1!)=10

Strings with one zero = 10!1!(10−1!)=10

Strings with two ones = 10!2!(10−2!)=45

Strings with two zeros= 10!2!(10−2!)=45.

1024 – 1 – 1 – 10 – 10 – 45 – 45 = 912.

How many bit strings of length 10 either begin with a bit 100 or stop with 11?

The solution is that there are 64-bit strings of period 10 that begin with 11 and stop with 0.

How many bit strings of length 8 both start with a 1 bit or end with 2 bits 00?
Hence, the variety of bit strings of period eight as a way to both begin with 1 or stop with 00 is 160.

How many bit strings are there of length 10 or less?

Solution: From the query, it’s miles for the reason that strings of period 10 include at the least 3 1s and at the least 3 0s. The general period is = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024.

How many 10 digit bit strings are there?

The first digit may be any of 5 digits, however, the difference may be any of 10 digits, so the solution is 5 · 10 · 10 = 500.

How many bit strings of length 10 contain both 5 consecutive zeros of 5 consecutive ones?

= 6 ways. Therefore, the solution is 6. We also can discover the solution through the variety of locations wherein the 3 0’s may be located collectively when it comes to the alternative 5 bits: 1.

Recurrence Relation Example 4 | Bit strings of length n that contain three consecutive 0s.

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