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FAQsu s constitution worksheet pdf answer key
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Out of 6 teachers and 4 students, a committee of 5 is to be formed. How many such committees can be formed including at least one student?
There are a total of 10 persons (6 teachers + 4 students) from which a committee of 5 is to be formed. This can be done in 10C5 ways.The number of ways that a committee can be formed with no students in it is to select 5 teachers out of the pool of 6 teachers available. This can be done in 6C5 ways.Therefore the total number of ways that the committee can be formed with at least one student in it is 10C5 - 6C5.10C5 = 2526C5 = 6Therefore the required number of ways = 246
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Five-digit numbers are formed using the digits 1, 2, 3, 4, 5, 6. Out of these, how many are even?
720 Five-digit numbers will be formed using the digits 1,2, 3,4,5,6. Out of them 360 Five -digit numbers will be even. You can do the math-If we place the 2,4,6 at one’s place the digits formed will be even_ _ _ _ 2Now total no.of ways to arrange 4 numbers out of 5 numbers will be 5! Which is 120Similarly when 4,6 are placed at one’s place, total no.of arrangements are 120 and 120So this gives 360 arrangements.There will be 360 such arrangements which will be even.Thanks for reading my answer i hope you have understand.
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A committee of 5 is to formed out of 6 gents and 4 ladies. In how many ways can this be done when each committee may have at the most two ladies?
At most 2 ladies means either no lady, 1 lady or 2 lady.Case 1: 0 lady, 5 members from 6 gents can be selected in 6C5 = 6 ways.Case 2: 1 lady, 1 lady can be selected from 4 ladies in 4 ways. Corresponding to each lady selected, rest 4 members can be selected from 6 gents in 6C4 = 12 ways. So, total ways in Case 2 is 48.Case 3: 2 lady, 2 lady can be selected from 4 ladies in 4C2 = 6 ways. Corresponding to two ladies selected, rest 3 members can be selected from 6 gents in 6C3 = 20 ways. So, total ways in Case 3 is 120.Now each case is independent. So, total number of ways is 120 + 48 + 6 = 174.Answer : 174
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A team of 4 is to be constituted out of 5 girls and 6 boys. What is the probability that the team may have 3 girls?
The word probability can’t go with may have. It has to be either “exactly 3” girls or “3 girls”Exactly 3 girls means the team consist of 3 girls and 1 boy.3 girls means the team can have 3 girls with 1 boy or 4 girls and no boy.Exactly three girls5C3 is the no of ways you can select girls. 6C1 is the no of ways you can select a boy. Thus the no of ways you can select the desired team composition is 5C3 * 6C1 which is 10 * 6 or 60. This is your favorable case. The ways of making a team of 4 people out of 11 is 11C4 which is 330. This is total number of cases.Probability is the ratio of favorable cases to total cases. The required answer is 60/330 or 18.18%Three girlsHere the favorable cases would be 3 girls or 4 girls as already explained above. As solved, total cases of 3 girls being selected is 60. Total ways of selecting 4 girls out of 5 is 5C4 which is 5. Thus total favorable cases become 60+5 or 65Probability is 65/330 or 19.69%
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How should a petitioner fill in Part 5 Household Size of the form I-864 Affidavit of support if he would like to sponsor 2 principal immigrants at the same time? Each family has 4 members.
Each principal beneficiary (and their family) is petitioned with a separate I-130 petition, and each I-130 petition has a separate I-864 Affidavit of Support. Each family’s I-864 does not count the other family in the “family members” in Part 3 (note that it says “Do not include any relative listed on a separate visa petition.”).If the two I-864s are filed at the same time for the two families, then each family’s I-864’s household size (Part 5) would just count the number of people immigrating in that family, which is 4 (item 1), the petitioner (item 2), and the petitioner’s spouse (item 3), dependent children (item 4), and other tax dependents (item 5), if there are any. It would not count anyone from the other family.On the other hand, if one I-864 is filed for one family, and that family has already immigrated before the second I-864 is filed for the other family, then the first family’s members will need to be counted in Part 5 item 6 (people sponsored on Form I-864 who are now lawful permanent residents) for the second family’s I-864.
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A Storage tank has two inlet pipes, one capable of filling the tank in 4 hrs and the other in 5 hrs and 30 mins. The outlet is capable of emptying the tank in 6 hours. How long will it take to fill such an empty tank if all pipes are turned on?
Let V = volume of a full tankVolume in per hour:V/4hours + V/5.5hours =5.5V/22hours + 4V/22hours = 9.5V/22hoursVolume out per hour:V/6hoursVolume in minus volume out:9.5V/22hours - V/6hours =28.5V/66hours - 11V/66hours = 17.5V/66hoursNow since 1V = one tank full, you want to solve for hours per 1V66hours / 17.5V = 66/17.5 hours to fill the tankor3.77 hours to fill the tankor226 minutes to fill the tankor13,577 seconds to fill the tank.
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How many ways are there to make 5 out of just the numbers 1, 2, 3, 4, 6, 7 by adding, subtracting, multiplying and dividing?
“How many ways are there to make 5 out of just the numbers 1, 2, 3, 4, 6, 7 by adding, subtracting, multiplying and dividing?”An infinite number of ways, since you didn’t say how many times I was allowed to repeat a digit, or a limit to how long an expression I could make.First, pick any “5”; let’s use “4+1”.Now, let’s create a bunch of “zero”-patterns:(4+2–6), (3+4–7), (2+1–3), or even longer ones like (7+6–4–3–3–2–1)Next, add as many zero patterns as you like, in whatever order you like, until you get tired of doing that. Then, since you tire so easily, write (or have someone write for you) a computer program to keep appending the next random choice of zero pattern to your ever-growing expression line.You can keep doing this until the universe ends, so there is no limit to the number of ways.
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How many numbers each of 5 different digits can be formed from the digits 0,4,5,6,7 if the digits 4 and 5 are to be next to each other?
[math]\{0,4,5,6,7\}[/math]Number cannot start with [math]0[/math], So there are [math]4[/math] choices for the first digit, [math]4[/math] choices for the second (because now I can choose [math]0[/math] too), [math]3[/math] choices for the next, [math]2[/math] for the next and [math]1[/math] for the last digit.So there are a total of [math]4\times 4\times 3\times 2\times 1[/math] number of combinations.Group the [math](45)[/math] or [math](54)[/math]. We can put them into [math](1,2), (2,3), (3,4)[/math], or [math](4,5)[/math] where [math](a,b)[/math] denotes the digit’s place in the number. So there are [math]4[/math] ways to put [math]4[/math] and [math]5[/math] together.The required number of combinations [math]= \dfrac{4\times 4\times 3\times 2\times 1}{4}=24[/math]Please correct me if I’m wrong, I still struggle with combinatorics
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