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Nairaland Forum / Nairaland / General / Education / Nairaland Mathematics Clinic (493969 Views)
Mathematics Clinic / 2015 Cowbell Mathematics Champion, Akinkuowo Honoured By School. / Lead City University Clinic Welcomes First Ever Baby Since 10 Years Of Opening (2) (3) (4)
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Re: Nairaland Mathematics Clinic by Aybalance: 2:56pm On Jan 20, 2018 |
lordbeans:workings pls... |
Re: Nairaland Mathematics Clinic by Aybalance: 2:59pm On Jan 20, 2018 |
Also this...If x is a real number and x+11<0 Evaluate /x/divided by x.thanks |
Re: Nairaland Mathematics Clinic by Olarewajub: 6:50pm On Jan 20, 2018 |
Some one should please simplify.
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Re: Nairaland Mathematics Clinic by Kentnickole(m): 8:51pm On Jan 20, 2018 |
Aybalance:n=7 |
Re: Nairaland Mathematics Clinic by lordbeans(m): 11:58am On Jan 21, 2018 |
Aybalance: nP3=n!/(n-3)!=n(n-1)(n-2)(n-3)!/(n-3)!=n(n-1)(n-2). -------(1) nC4=n!/[(n-4)!4!]=n(n-1)(n-2)(n-3)(n-4)!/[(n-4)!4!]=n(n-1)(n-2)(n-3)/4!=n(n-1)(n-2)(n-3)/24. --------(2) (1)+(2)=6; => n(n-1)(n-2)+n(n-1)(n-2)(n-3)/24=6. ----------(3). Multiply both sides of (3) by 24. =>24 n(n-1)(n-2)+n(n-1)(n-2)(n-3)=144.----------(4). =>factorising (4) we have n(n-1)(n-2)(21+n)=144.......(5). from (5) one can see that n=3 satisfies the equation. Hence the solution is n=3. |
Re: Nairaland Mathematics Clinic by lordbeans(m): 12:01pm On Jan 21, 2018 |
Aybalance: -1 |
Re: Nairaland Mathematics Clinic by lordbeans(m): 12:16pm On Jan 21, 2018 |
Re: Nairaland Mathematics Clinic by Olarewajub: 12:37pm On Jan 21, 2018 |
lordbeans:Please can you screenshot the solution. Will really appreciate. Happy Sunday |
Re: Nairaland Mathematics Clinic by UTILITYMAY(m): 10:09pm On Jan 21, 2018 |
Please guyz... I need solution to question 10 1 Like
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Re: Nairaland Mathematics Clinic by Aybalance: 8:47am On Jan 22, 2018 |
Kentnickole:pls can u show the workings...bcus the lord of beans got 3...i really need to clarify |
Re: Nairaland Mathematics Clinic by Aybalance: 8:48am On Jan 22, 2018 |
lordbeans:pls the workings.thanks |
Re: Nairaland Mathematics Clinic by Mechanics96(m): 12:13am On Jan 23, 2018 |
lordbeans: N cannot be 3 as 3C4 will never exist. You don't expect us to select 4 items out of 3, do u? That was a wrong question, it should be Np3÷Nc4 =6, and n=7 |
Re: Nairaland Mathematics Clinic by Mechanics96(m): 12:24am On Jan 23, 2018 |
lordbeans: Wrong sir, your answer '3' violates the rules of distribution... 3c4 will never exist. The question is wrong in the first place! The question is Np3÷NC4=6 and n=7 N! /(n-3)! × (n-4)!4!/n! =6 (n-4)!4!/(n-3)(n-4)! =6 4!/(n-3) =6 6(n-3) =4! 6n-18=24 N=42/6 N=7 |
Re: Nairaland Mathematics Clinic by lordbeans(m): 5:02am On Jan 23, 2018 |
Mechanics96:It is true that nCr only exist when n>=r >=0 (in other to avoid negative factorial). However, if you narrow down on the meaning of this formula, you quickly realize that it just means how many number of ways in which a given 'r' quantity can be gotten from an 'n' quantity (e.g. 8C5 means how many ways can you choose 5 eggs given 8 eggs). In this scenario 3C4 means how many number of ways can you choose 4 eggs given 3 eggs---the logical answer is 0 ways (because you cannot). So in as much as n>=r for the above stated formula to exist, the converse can indeed be logically derived in the same way 0^0 does not exist (this can be proved) but for convenience some real analysis textbooks assume 0^0=1 (the good ones state this assumption in the footnotes). PS: n=7 is only correct if the OP admits the question is erroneous and it is the way you have stated it. |
Re: Nairaland Mathematics Clinic by ProjectShelve(m): 7:44am On Jan 23, 2018 |
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Re: Nairaland Mathematics Clinic by Mechanics96(m): 6:48pm On Jan 23, 2018 |
UTILITYMAY: M=4 and m= - 1. Bn trying to upload the solved piece but keep failing |
Re: Nairaland Mathematics Clinic by UTILITYMAY(m): 7:44pm On Jan 23, 2018 |
Mechanics96: Thanks I solved it that same night; 1 Like
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Re: Nairaland Mathematics Clinic by UTILITYMAY(m): 7:44pm On Jan 23, 2018 |
Mechanics96: Thanks I solved it that same night;
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Re: Nairaland Mathematics Clinic by Kentnickole(m): 11:05pm On Jan 23, 2018 |
Mechanics96:Aybalance sori to hv replied late. Very busy, pardon me. Mechanics96 has done it all. 1 Like |
Re: Nairaland Mathematics Clinic by Aybalance: 1:53pm On Jan 28, 2018 |
Kentnickole:No problems sir...seems there was an error from my book...thanks |
Re: Nairaland Mathematics Clinic by Xsem(m): 3:20pm On Jan 28, 2018 |
Mathemen help
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Re: Nairaland Mathematics Clinic by Xsem(m): 10:58pm On Jan 28, 2018 |
What is the difference among all of them please.
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Re: Nairaland Mathematics Clinic by Missboyy: 9:25am On Jan 31, 2018 |
Xsem: (a+b) 2 = a2 = 2ab = b2 Open the brackets: (a+b)(a+b) [+ signs in both brackets gives all postives] (a-b)2 = a2 -2ab + b2 [- sign in both brackets. The number/letter after the sign will be positive because multiplying a negative by a negative gives a positive] (a+b)(a-b) [ one - sign and one + sign] That's the difference of 2 squares. (a+b)(a-b) = a2 - ab + ab -b2 = a2 -b2 So a2 -b2 = (a+b)(a-b) NOT (a-b)2 (see that answer above) and a2 + b2 I don't think you can expand that any further. That means that a2 + b2 does not equal (a+b)2 (See the first point). So.. (a+b)2 = (a + b)(a - b) = a2 +2ab +b2 (a-b)2 = (a - b)(a - b) = a2 -2ab +b2 a2 - b2 = (a + b)(a - b) AKA Difference of 2 squares! a2 +b2 = a2 + b2 |
Re: Nairaland Mathematics Clinic by Xsem(m): 12:58pm On Jan 31, 2018 |
Missboyy:Thank you though there are other ones I posted along with this one. |
Re: Nairaland Mathematics Clinic by Mechanics96(m): 3:14pm On Jan 31, 2018 |
weldone bro, let me just add that a²+b² =(a+b)² - 2ab It is called sum of two squares...see algebraic processes in your mathematics textbook for more. |
Re: Nairaland Mathematics Clinic by NelsonObas(m): 10:30pm On Feb 01, 2018 |
please, some one help on the no. 6 |
Re: Nairaland Mathematics Clinic by NelsonObas(m): 10:33pm On Feb 01, 2018 |
NelsonObas:
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Re: Nairaland Mathematics Clinic by Aybalance: 8:47am On Feb 02, 2018 |
Pls i need help on this...A photograph is to be taken of five people including married couple.If the married couple insist on sitting next to each other,the number of different arrangement is? |
Re: Nairaland Mathematics Clinic by Mechanics96(m): 9:08am On Feb 02, 2018 |
[quote author=NelsonObas post=64707346][/quote]
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Re: Nairaland Mathematics Clinic by Mechanics96(m): 10:27am On Feb 02, 2018 |
Aybalance: The 2 people(couple) are taken as 1, since they cannot be separated in the course of the arrangement. Therefore, (5-1) people are to be arranged. That is 4! But the two people can be interchanged ( arranged) in 2! Ways That makes it 4! × 2! = 48ways |
Re: Nairaland Mathematics Clinic by NelsonObas(m): 10:28am On Feb 02, 2018 |
[quote author=Mechanics96 post=64714751][/quote]the (ii) |
Re: Nairaland Mathematics Clinic by lordbeans(m): 10:58am On Feb 02, 2018 |
Aybalance: Let us assume the married couple to be CD and the rest people to be X. So the possible arrangements are CDXXX XCDXX XXCDX XXXCD. 4 arrangement of which in each arrangement you have 3!(6) ways . In total 24 ways. But you can also swap the couple so that D comes before C DCXXX XDCXX XXDCX XXXDC. which is another 24 ways, making the total to be 48 ways. Ans 48 ways. |
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