4/5 of the 315 members of a book club are male. How many female members are there in the club?
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Female mems = 315 * 1/5 = 315:5=63\n\n63 is the final answer
1034525
Jimbo is the first person with the correct answer.\n\nThe first book you put on the shelf can be selected from the 6 books available and can be placed on the shelf either upright or upside down.  There are 12 possibilities for the first book (6 books multipled by two ways round).\n\nWith one book on the shelf you now have 5 books remaining from which you select one and place it either upright or upside down on the shelf.  But now there are only 10 possibilities (5 x 2).\n\nWith 4 books remaining there are 8 possibilities (4 x 2).\nWith 3 books remaining there are 6 possibilities (3 x 2).\nWith 2 books remaining there are 4 possibilities (2 x 2).\nWith 1 book remaining there are 2 possibilities (1 x 2).\n\nTo calculate the total number of possible ways to arrange the books you need to multiply together all the individual possibilities... 12 x 10 x 8 x 6 x 4 x 2... which equals 46,080.
236699
For the first question, that means you want to know about committees with exactly 3 women and committees with exactly 4 women. Since these are independent cases, we consider each separately and add the result. \nFor 3 women, we need to know how many unordered groups of 3 women we can choose from 4 women, so we have 4[Choose]3. The men have 2 people on this committee, so they contribute 4[Choose]2 to the product for a total of (4[Choose]3)*(4[Choose]2) committees with exactly 3 women on board. \nThere is only one way to grab an unordered set of 4 objects from a set of 4 objects, so we go right to the men. There are 4 ways the men can fill the last spot, so this is 1*4=4 different committees with 4 women on board.\nThus the total number of committees with a majority of women on board is (4[Choose]3)*(4[Choose]2) + 4.\nThe second part is very similar to the first.
734577
First, find out how many ways you can arrange 20 different books with exactly two books on each shelf.  Then, find how many ways you can arrange the remaining 10 books on the 5 shelves.  Multiply these two answers together to find the total number of arrangements.  \n\nFor the first, consider each spot on each shelf as a different position.  How many choices are there for the left-most spot on the top shelf?  How many choices for the second spot on the top shelf?  Continue for each shelf and multiply them together - you should get 20!/10!\n\nFor the second, you have 10 books and 5 shelves where each book can go.  A shelf can have no books at all or it might have all 10 books.  If you start at the top shelf, they can either place a book or move to the next shelf.  You have 10 books and will move to the next shelf a total of 4 times.  Pretend you are putting the books in a row and using a marker to indicate when you move to the next shelf - you have 14 spots to place either a book or a marker and the markers all look alike.  (This is similar to arranging the letters AAAABCDEFGHIJK)  The number of ways you can arrange them is 14! / 4!\n\nThus, the total number of ways to arrange the books is 20!14! / 10!4!
1011403
1,257,473... HA HA HA HA HA
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69..\n\nand try to paraphrase your friggin homework...
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each child can be a male or a female \ntherefore the total no of events possible is 2^6=64\n\nthe question says that you should have atleast 3 girls \ni.e there can be three, four ,five or all six girls .\n\nthe number of ways of choosing 3,4,5,6 girls from 6 children is the sum\n6C3 + 6C4 + 6C5 + 6C6 (here C stands for combinations )\n\ni,e\n\n(6*5*4)/(3*2*1) + (6*5*4*3)/(4*3*2*1) + (6*5*4*3*2)/(5*4*3*2*1) + 1\nthis reduces to \n20+15+6+1\n=42\n\ntherefore the probability is \n42/64
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Your question is not politically correct!\n\nTheir are 365 men in the House Of Representatives and 70 women. \n\nTheir are 86 men in the Senate and 14 women.\n\nCount the reps from each state here:http://www.house.gov/house/MemStateSearch.shtml
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all of them are girls...\n\nso half of them are girls, so is the other half
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There are (10 choose 3) = 120 possible committees that can be created.  To get one of all women, select one of the six women first, one of the five left next, and then one of four.  The order you choose should not matter so divide by 3!.  So there are 6*5*4/3! = 20 possible all women committees.\n\n20/120 = 1/6
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5 boys, 2 girls \n\n(D has one sister, and 6 brothers, Tom has 2 sisters, and five brothers)\n\nfive brothers + Tom\n1 sister + D
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This would be a binomial.  Assuming that p=1/2 , equal probability of male or female.\n\nThen it's like 80 trials with success (male) or failure (female)\n\nAt least 30 male means that you have to add probs of 30, 31, 32, ...,80.  It's easier to add probs of 1,2,...,29 and subtract from 1.\n\n80 is a large enough number to use the normal approximation.\n\nFor the normal, the mean would be 80/2 = 40.  So, at least 30 is\n\n1-P(X<less than 30) \n\nconvert X to Z to use standard tables.
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she could have 4, 5, 6, or 7 boys. in any of those cases, at least half are boys.
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Women now make up 56 percent of the college population.\n\nI found it here:
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One can have only one day with exactly 129 books, because\n600-2*129 = 342 < 365-2\nand each day at least a book was sold.
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Forty of course.\n\n=D
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Let x be the number of girls.\nx+x3/5=40\n 5x+3x=200\n       8x=200\n         x=200/8\n         x=25\nSo the number of girls is 25.\nAnd the number of guys is 25*3/5=15.