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Get And Sign Export Certification Forms 9060 6 And 9060 5 1 Form 9060 6 Is The Aglearn Usda
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FAQs aglearn usda gov
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A committee of 5 persons is to be formed from 6 male and 5 female members. In how many ways this committee can be formed so that there are at least 1 male and 1 female member in the committee?
We could count out every single combination of “x” males and “y” females that satisfied the condition (e.g. 1 male, 1 female; 2 males, 1 female; etc). But, it’s a lot easier in this case to count the number of ways where there aren’t at least 1 male or 1 female in the committee.Case 1: There are no females in the committee.In that case, you have 6 males to choose from to go on your 5-person committee. Assuming that the order in which you pick the members of the committee doesn’t matter, we can say there are “6 choose 5” ways* for the guys to go on this committee. “6 choose 5” = 6!/(5! × (6 - 5)!), or just 6.Case 2: There are no males in the committee.In that case, you have 5 females to choose from to go on your 5-person committee. Obviously, there’s only 1 way you can do that.So, there are only 7 ways you can choose the committee members where every members is of the same gender. Now, we have to count the total number of possible committees. There are 11 total members and 5 committee positions, so there are “11 choose 5” possible ways to order the committee. “11 choose 5” is just 11!/(5! × (6 - 5)!), or 462.So, there are 462 possible ways to order this committe and only 7 fail to meet the condition. Thus, there are 462 - 7, or 455 ways to order the committee.*“choose” is the word used to describe combinations. “n choose r” is the # of ways r things can be chosen from n things, without caring about the order in which the r things are chosen. “n choose r” = n!(r! × (n - r)!). -
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 -
How do I solve this question? A committee of 5 men and 3 women is to be formed out of 7 men and 6 women. If two particular women are not to be together in the committee, what is the number of committees formed?
Total Available Persons = 7 Men and 6 WomenA committee of 5 Men and 3 Women is to be formed.So, I need to select 5 men out of 7 and it can be done in [math]^7c_5[/math] ways3 women out of 6 can be selected in [math]^6c_3 [/math]waysNow, without any constraints, the number of ways the committee can be selected is:[math]^7c_5 \times ^6c_3 = 21 \times 20 = 420[/math] waysBut, the question says two particular women should not be there in the committee together. So, to find that, we need to look at the question in another way.Number of ways in which two particular women aren't there together in the committee is same as total number of ways minus the number of ways those two women are definitely there together.Number of ways in which the two women are definitely there can be calculated as:[math]^7c_5 \times ^4c_1 = 21 \times 4 = 84[/math]Note: As the two women are definitely there, so I just need to select one more woman from the remaining 4.Thus, number of ways to select such that two particular women are never there together is [math](420 - 84) = \fbox {336 (Answer)}[/math] -
A three digit number is to be formed using the digits {1,2,3,4,5,6}. How many numbers are possible if the digits can be repeated and the last digit must be odd?
well we know that the number is of the form xyzThe z can have 3 values, x and y can have any 6, sowe can say that for any xy we have 3 possible out comesxy1xy3xy5xy have 6*6 possibilities that’s 36 and 36*3=108. here are all these numbers:111113115121123125131133135141143145151153155161163165211213215221223225231233235241243245251253255261263265311313315321323325331333335341343345351353355361363365411413415421423425431433435441443445451453455461463465511513515521523525531533535541543545551553555561563565611613615621623625631633635641643645651653655661663665Yepp the count checks out.
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