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Re: Nairaland Mathematics Clinic by Aybalance: 2:56pm On Jan 20, 2018
lordbeans:

n=3
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.

Re: Nairaland Mathematics Clinic by Kentnickole(m): 8:51pm On Jan 20, 2018
Aybalance:
pls i need help on this... ........nP3+nC4=6.find n
n=7
Re: Nairaland Mathematics Clinic by lordbeans(m): 11:58am On Jan 21, 2018
Aybalance:
workings pls...

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:
Also this...If x is a real number and x+11<0 Evaluate /x/divided by x.thanks

-1
Re: Nairaland Mathematics Clinic by lordbeans(m): 12:16pm On Jan 21, 2018
Olarewajub:
Some one should please simplify.


2(3)5-2n
Re: Nairaland Mathematics Clinic by Olarewajub: 12:37pm On Jan 21, 2018
lordbeans:


2(3)5-2n
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

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Re: Nairaland Mathematics Clinic by Aybalance: 8:47am On Jan 22, 2018
Kentnickole:
n=7
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:

-1
pls the workings.thanks
Re: Nairaland Mathematics Clinic by Mechanics96(m): 12:13am On Jan 23, 2018
lordbeans:


n=3

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:


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.

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:


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
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:
Please guyz... I need solution to question 10

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:


M=4 and m= - 1. Bn trying to upload the solved piece but keep failing

Thanks I solved it that same night;

1 Like

Re: Nairaland Mathematics Clinic by UTILITYMAY(m): 7:44pm On Jan 23, 2018
Mechanics96:


M=4 and m= - 1. Bn trying to upload the solved piece but keep failing

Thanks I solved it that same night;

Re: Nairaland Mathematics Clinic by Kentnickole(m): 11:05pm On Jan 23, 2018
Mechanics96:


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
Aybalance sori to hv replied late. Very busy, pardon me. Mechanics96 has done it all.

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Re: Nairaland Mathematics Clinic by Aybalance: 1:53pm On Jan 28, 2018
Kentnickole:
Aybalance sori to hv replied late. Very busy, pardon me. Mechanics96 has done it all.
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

Re: Nairaland Mathematics Clinic by Xsem(m): 10:58pm On Jan 28, 2018
What is the difference among all of them please.

Re: Nairaland Mathematics Clinic by Missboyy: 9:25am On Jan 31, 2018
Xsem:
What is the difference among all of them please.

(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:


(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
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:
please, some one help on the no. 6

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]

Re: Nairaland Mathematics Clinic by Mechanics96(m): 10:27am On Feb 02, 2018
Aybalance:
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?


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:
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?

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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