(Permutations) Solve for n. (n)P(3) = 210?
Basically you are dealing with the following:n! / (n-3)! = 210We know that n! = n * (n-1) * (n-2) * (n-3) * .... * 1Which, for our use, could better be stated as n! = n * (n-1) * (n-2) * (n-3)!Plug this in:n * (n-1) * (n-2) * (n-3)! / (n-3)! = 210n * (n-1) * (n-2) = 210I would probably start to guess and check here, but if you want, you can multiply them all out:n * (n^2 - 3n +2) = 210n^3 - 3n^2 + 2n = 210Personally, I don't see how that would help you much. But either way you solve it, you'll get n=77 * (7-1) * (7-2)7 * 6 * 5 = 210210 = 2107^3 - 3(7^2) + 2(7) = 210343 - 147 + 14 = 210210 = 210Hopefully that helps you out! Factorials are tricky to simplify!
Permutations Question Solve for n?
3 * (n + 1)! / (n - 1)! = 1263 * (n + 1) * n * (n - 1)! / (n - 1)! = 1263 * n * (n + 1) = 126n * (n + 1) = 42n^2 + n = 42n^2 + n - 42 = 0n = (-1 +/- sqrt(1 + 168)) / 2n = (-1 +/- 13) / 2n = -14/2 , 12/2n = -7 , 6n = 6
Permutation equation?
Permutations of placing 5 objects in 11 bags, order important. I assume when you said randomly, they place only 0 or 1 items in each bag (not 2-5 in the same bag).Call the objects A,B,C,D,E[We do not need to consider using 11 symbols '/' for skipping a bag, since we said they place 0 or 1 items in each bag.]I actually disagree with your assertion that the order the gifts are placed is important, since all that should matter is whether gift X ended up in bag Y, not in which turn it was placed there.Hence I think you should be considering combinationsn!/r!(n-r)!Thus, combinations of 5 objects in 11 bags.No. of total combinations:(n!)/r!(n-r)! = 11!/5!(11-5)! = (11!/5!6!) = 462There is clearly only one 5-out-of-5 correct permutation.(Assume WLOG it was A,B,C,D,E in the first 5 bags.)P(5-out-of-5 placed correctly) = 1 / 462How many 4-out-of-5 correct permutations?Well, this involves placing any 4 objects correctly and 1 incorrectly.- There is only one way to place the 4 correct objects- For the 1 incorrect object, there are C(5,1)=5 ways to choose which object gets placed incorrectly, and there are C(11-4-1,1) = 6 eligible incorrect bags to place it in.P(4-out-of-5 placed correctly) = 1*C(5,1)*C(11-4-1,1) / 462= 30/462= 5/77 in lowest termsHow many 3-out-of-5 correct permutations?i.e. placing 3 of the 5 objects correctly and 2 of the 5 incorrectly.- There is only one way to place the 3 correct objects- There are C(5,2)=10 ways to choose the 2 incorrectly-placed objects, and there are C(11-3-1,2) = C(7,2) = 21 eligible incorrect bags to place it in.P(3-out-of-5 placed correctly) = 1*C(5,2)*C(11-3-1,2) / 462= 10*21 /462= 5/11 in lowest termsAnd in general for the rest,P(k-out-of-5 placed correctly)- k objects placed correctly and (5-k) placed incorrectly- There is only one way to place the k correct objects- There are C(5,5-k) ways to choose the (5-k) incorrectly-placed objects, and there are C(11-(5-k)-1,2) = C(6+k,k) eligible incorrect bags to place it in.=>P(k-out-of-5 placed correctly) = C(5,5-k)*C(6+k,k)/ 462[And the rest is just button-bashing.]
Solving permutation equations?
The solution is n=10. To see this, you need to realize that n!/(n-6)!=n(n-1)(n-2)(n-3)(n-4)(n-5), and that n!/(n-5)!=n(n-1)(n-2)(n-3)(n-4). Then your equation becomesn(n-1)(n-2)(n-3)(n-4)(n-5)=5*n(n-1)(n-...then all the factors n(n-1)(n-2)(n-3)(n-4) cancel out of both sides, and you are left withn-5=5son=10
Solve for n: nP3 = 24 (permutation)?
nP3 = n!/(n-3)! = n(n-1)(n-2) so we want n such that n(n-1)(n-2) = 24. We could solve this the hard way but since n is an integer just try a few numbers to find that n=4 satisfies the equation.
2 Perms and combs questions?
P(4,4) = 4*3*2*1 = 24P(4,3) = 4*3*2 = 24The interpretation of these is: 4 letters for first spot, 3 remaining letters for second spot and so on. (below is an illustration of our empty spots, the first being the far left)_ _ _ _The *1 at the end of P(4,4) is the number of letters remaining to fill the final spot, which is 1 since there are the same number of letters as spots to fill.Hence for each of the ways of arranging 3 letters from 4 the final unused letter can take the last spot in precisely 1 way.In other words: for every permutation of 3 letters from 4, a permutation of 4 letters from 4 can be created by *appending* the final unused letter. Conversely every permutation of 3 letters from 4 can be created by *removing* the final letter from every permutation of 4 letters from 4. This one-to-one correspondence is known as a bijection.The first few maps for this bijection are below:ABCD <-> ABCABDC <-> ABDACBD <-> ACB...P(N,3) = 2C(N,4)N(N-1)(N-2) = 2N(N-1)(N-2)(N-3)/4!4! =2(N-3)12 = N-3N = 15
Permutation help! nP4 = 42(nP2)?
Formula 1: n P k = n! / (n-k)!Formula 2: n! = n. (n-1)!Therefore, n P 4 = n! / (n-4)!Problem:n P 4 = 42 (n P 2)=> n! / (n-4)! = 42. n! / (n-2)! [using formula 1 from above]=> 1/ (n-4)! = 42 /(n-2)! [dividing both sides by n!]=> (n-2)! = 42. (n-4)! [multiplying both sides by (n-2)! * (n-4)! ]=> (n-2)(n -3) (n-4)! = 42. (n-4)! [using formula 1 from above]=> (n-2)(n-3) = 42 [dividing both sides by (n-4)!]=> n^2 -2n -3n +6 -42 = 0=> n^2 - 5n - 36 = 0Now, we need to find factors, who sum = -5Factors from x^2 term and independent term = -36 = (-1)* 2 * 2 * 3 * 3If we choose (2*2) & (-3*3), their sum = -9 + 4 = -5 (accepted)Verify: if we choose 2 & -2*3*3, subtraction = -18+-2 = -16 (Not accepted) Xif we choose 3 & -2*2*3, subtraction = -12 +3 = -9 (Not accepted) X and similar.=> n^2 - 9n + 4n - 36 = 0=> n(n-9) + 4 (n-9) = 0=> (n-9) (n+4) = 0so, (n-9) = 0 or (n+4) = 0=> n = 9 or n =-4n can't be -ve, so n=-4 is not accepted.Therefore, n = 9Answer: n = 9Verify:n P 4 = 42 (n P 2)=> 9 P 4 = 42. 9 P 2=> 9! /(9-4)! = 42. 9! /(9-2)!=> 9! / 5! = 42. 9! / 7!=> 9!/5! = 7. 6. 9! / (7.6. 5!)=> 9!/5! = {(7.6)/(7.6)} . 9!/5!
Permutations and Combinations Math help?
4.After everyone had shaken hands once with everyone else in a room, there was a total of 21 handshakes. How many people were in the room?Each handshake represents one possible combination of two people from the number (call it n) in the room, and the total number of combinationsnC2 = 21n! / [(n-2)! 2!] = n (n-1) / 2 = 21n^2 - n = 42n^2 - n - 42 = 0(n - 7) (n + 6) = 0and since a solution of -6 is no use when counting people, n=7.5. Solve each of the following for n. Put the values of n in order, from smallest to largest.This is just more solutions like the previous one. (I'll leave you to put the answers in order.)1) nC1 = 18n! / [(n-1)! 1!] = 18n = 182) 8Pn = 3368! / (8 - n)! = 3367! / (8 - n)! = 336/8 = 426! / (8 - n)! = 42/7 = 65! / (8 - n)! = 6/6 = 15! = (8 - n)!5 = 8 - nn = 33) nC3 = 3(nP2)n! / [(n-3)! 3!] = 3 n! / (n-2)!n! (n-2)! / (n-3)! = 3 n! 3!(n-2)! / (n-3)! = 18n - 2 = 18n = 204) nP3 = 2(nC4)n! / (n-3)! = 2 n! / [(n-4)! 4!]4! n! = 2 n! (n-3)! / (n-4)!12 = (n-3)! / (n-4)! = n - 3n = 156. How many different bridge hands are possible, if a bridge hand consists of 13 cards dealt from a standard deck of 52 cards?52C13 = 52! / (39! 13!) = 635,013,559,600= approximately answer d
Please help me solve: 6P3?
It stands for "permutation". When you see "n P r", it means the same asn! / (n-r)!So 6P3 is6! / (6-3)! =6! / 3! =6*5*4 =120This is the number of permutations you can make with 6 objects, taking 3 at a time. So for example if you had the letters "a, b, c, d, e, f" and wanted to know how many ways you could make groups of 3 with them (like "abc", "adf", "cef", etc.) there would be 120 of them.
I have a problem with this math question....help!?
let p= the number of pens and n= the number of notebooks. Then:3p + n = 9 andp + 2n = 8First, solve one of the equations for a variable. Let's solve the top in terms of n. Then we get n = 9 - 3p. Substitute that into the second equation for n:p + 2(9 - 3p) = 8 Now, solve for p:p + 18 - 6p = 8-5p = -10p=2. Now throw 2 in for p in either equation:2 + 2n = 8. Solve for n2n=6n=3. Throw these into the top equation and see if it checks:3(2) + 3 = 99=9It works! So pen's are 2 dollars and and notebooks are 3 dollars.Hope it helps!