For each problem, display the equation used to form the answer. Leave combinatio

Question

For each problem, display the equation used to form the answer. Leave combinations, permutations, factorials, and exponents intact.
7. (2 points) How many bit strings contain exactly five Os and 14 Is if every 0 must be immediately followed by at least two 1s? 8.(1 point) In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if exactly one of the bride and the groom is in the picture? 9.(I point) How many bit strings of length seven either begin with two Os or end with three 1s?

Explanation / Answer

7) As mentioned each 0 must be followed by atleast two 1s.

i.e if first element is 0 then atleast next two elements must be 1s ex 011.

as mentioned we can have max 5 zeroes, so considering 011 we can have 5 sets of 011 and then 4 1s to fill the condition.

Now to calculate the number of ways it can be done we have to use permutations and combinations.

Calculating this we would get the eanswer 126.

8) We are given that the picture has to be taken for 6 people out of 10 where exaclty one of the bride or the groom has to be present in the picture.

since either of them has to be present in the picture, we have to select 5 out of 9 people.

So first, we calculate the number of ways in which 6 people can stand together, that will be 6!

Now since one of them is either the bride or the groom we have to select 5 people of the remaining 9 people but out of them one would be the groom or bride accordingly which cant be in the picture so that leaves us to 5 out of 8 people to choose from.

So now we have to calculate 8 choose 5 ie 8c5 which would give us 56 ways to choose 5 people out of 8 people.

Now since we have selected 6 people we can arrange them in 6! possible ways, i.e. in 720 ways.

So choosing 6 people and arranging them in different ways is calulated by 720*56 which would give 403202 ways

Now this was case for one of them, So to calculate this for both of them we would have to multiply it by 2 which would give us 80640 ways.

9) Here we need to calculate the number of ways in which the string of length seven either begin with two 0's or end with three ones.

So for the fist case begining with 2 zeros we have to calculate how the remaining 5 digits can be rearranged and for 2nd case we need to check how the first 4 elements can be rearranged.

So for the first case we can arrange them in 2*2*2*2*2 ways, i.e 32 ways

for the second case we can arrange them in 2*2*2*2 ways i.e. 16 ways

so total being 48, however we are double calculating few scenarios here, like the string begining with double 0 ending with tripple 1's .

for this we can also use the formula based on inclusion-exclusion principle which would give us the answer.

2 RaiseTo(72) + 2RaiseTo( 73) 2RaiseTo( 75 )= 44

so here 44 is the final answer.

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