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# Math help show me the steps to solve these problems?

Topic: Five steps for problem solving
June 25, 2019 / By Kelley
Question: In how many ways can a president, a vice-president, and a secretary be selected from an organization of 33 members? A club has 26 members and 11 pledges. A president, vice-president, and secretary are selected from the members, and a pledge chairman and pledge vice-chairman are selected from the pledges. In how many ways can these five officers be selected? How many different "words" are possible using all the letters of ABELITE? How many different color arrangements are possible by placing 1 green ball, 1 red ball, 1 yellow ball, and 4 tan balls in a row?

## Best Answers: Math help show me the steps to solve these problems?

Herman | 3 days ago
for the ABELITE one, im pretty sure you count the letters, so theres 7 letters. and you multiply, i remeber learning something like this a long time ago, but i forget what it's called. i think you multiply 1 x 2 x 3 x 4 x 5 x 6 x 7 and that would be 5040. im sorry, i learned this so long ago so i forget a lot of it but i think i'm right about the abelite question. good luck! (:
👍 254 | 👎 3
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Originally Answered: Math help show me the steps to solve these problems?
for the ABELITE one, im pretty sure you count the letters, so theres 7 letters. and you multiply, i remeber learning something like this a long time ago, but i forget what it's called. i think you multiply 1 x 2 x 3 x 4 x 5 x 6 x 7 and that would be 5040. im sorry, i learned this so long ago so i forget a lot of it but i think i'm right about the abelite question. good luck! (:

Ellgar
I hated this unit in high school 33 members. 3 officer positions. Officer positions are distinct - it matters which of the officers is in which position. Because the offices are distinct, this is a permutation. 33 P 3. Check/explanation: 33 members. Any one of them can be the president. That's 33 possibilities. 32 remaining members, and any one of them can be vp. That's 32 possibilities for each of the 33 choices for president, because the combination of president and vp will be different each time. So 33*32. 31 members remain, any of them can be secretary, and for the same reasoning as before, that means 31 combinations for each of the 33*32 combinations already in the works. So 33*32*31. And that's the definition of 33 P 3. For the same reasons as above, 26 P 3 for the officers out o fthe membership. 26*25*24. And for the same reason, 11 P 2 for the pledge officers. And for the same reason, because the overall lineup of 5 officers with specific positions differs with every rearrangement and different choice, you have to multiply the choices for pledge officers by the choices for member officers (26 P 3) * (11 P 2) This is a factorial. The order of the letters is what is in question, so clearly it uses the permutation function rather than the combination function. And since you need to use all 7 letters in all of your combinations, it would be 7 P 7. That's the definition of 7! . This is messy. The basic form is still 7! : you have 7 balls and you need to use each of them in every combination. However, it isn't a pure factorial, because 4 of the balls are identical. Identical balls can be interchanged, which means that even if an arrangement assigns each tan ball individually to a different location, it would be repetitive (and thus not a unique arrangement) with an arrangement that put different tan balls in the same locations. So you have to divide by the number of times you can make those switches, which is 4! So the answer is 7! / 4! = 7 P 3
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Christian
I assume that a person cannot hold more than one office. This is "selection without replacement." First Problem: 33 choices for president times 32 choices for vice-president times 31 choices for secretary = 33*32*31 = 32,736 possibilities. Second Problem: 26 choices for president times 25 choices for vice-president times 24 choices for secretary times 11 choices for chairman times 10 choices for vice-chairman = 26 * 25 * 24 * 11 * 10 = 1,716,000 possibilities. Third Problem: This is another selection without replacement, but with some objects that appear more than once (the E). We have seven letters, so we have 7! words, but we have two letter E's, so we divide this by 2 to get the answer: 2,520. In this case it is easy to handle the repeated letter. In general it is more complicated. If we had had three E's, then we'd have 3! ways to arrange these "different" E's, or 6. We'd have to divide by 6 instead of 2. See the next problem. Fourth Problem: Seven balls gives 5,040 possibilities. Now we have 4 tan balls, and they can be arranged in 4! = 24 different ways. Thus we divide 5,050 by 24 to get 210 different arrangements.
👍 108 | 👎 -9

Ambrose
1. 33P3 = 33*32*31 = 32736 2 .26P3 * 11P2 = ( 26*25*24 ) (11*10) = 1716000 3. Using all the letters it will be 7P7 = 7*6*5*4*3*2*1 = 5040 4. Bit tricky, Since 2 or more tan balls together, is not considered as separate color arrangement, Consider differnet cases of tan balls together, 4P4 (assuming all the 4 tan balls are together )+ 3P3(3 tan balls together)*4P4+ ........................... I give up!!
👍 106 | 👎 -15

Tikva
marbles probability green tan green yellow green red red yellow red tan yellow tan tan tan or: green tan green red green yellow red tan red green red yellow yellow red yellow green \yellow tan tan green tan yellow tan red tan tan.. i know that green red and like red green are technically the same thing.. so if you dont want the repition of colours then the first one is probably correct
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Originally Answered: How do you solve SAT function problems?
For all positive values of p, let p*smiley face symbol* be defined as p/(p^2+1). I've made two assumptions here: (1) that your changing p from lower- to upper-case was a mistake; and (2) that the +1 belongs in the denominator of the fraction, although you failed to indicate that properly. What is the value of 2*smiley face symbol*. Under those assumptions, the point here is that we replace p with 2: 2*smiley face symbol* = 2/(2^2+1) = 2/5 In general, book advice contains assumptions about the form of the problem that might not apply to a particular case. In the general case described by the book, the problem could have been presented this way: f(x) = x / (x^2+1) What is the value of f(2)? The use of p instead of x, and of the smiley face symbol instead of the usual function notation f(x), is a different expression of the concept. You need to learn to look past such differences to discern what's really being asked.

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