Isn't the formula this?
For the number of ways to arrange n objects taken r at at a time
n!/(n-r)!
So it would be
7!/(7-4)! = 7!/3! = 7*6*5*4*3!/3! = 7*6*5*4=840
But since we have 2 c's and 2 o's, the answer would be 840/2!2! = 840/4 =210
Last edited by Mongoose528; 3 Aug 2016 at 10:03 AM.
How do I do this
Also
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anyone who can help me with question 12b and 13a of exercise 4F?
Last edited by Drongoski; 27 Sep 2016 at 11:02 PM.
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The first question:
The "sum of the products of pairs of roots" refers to the x term which is absent, meaning the coefficient is 0. Therefore the "sum of the products of pairs of roots" = 0
Algebraically: ab + a(a+b) + b(a+b) = 0 [... a = alpha, b = beta]
Expanding: ab + a^2 + ab + ab + b^2 = 0
ab + a^2 + 2ab + b^2 = 0
Combining the perfect square:
ab + (a+b)^2 = 0
And you can use the results from the question before 12a what (a+b)^2 is.
That is substitute (a+b) = 1/2*A = A/2
ab + (A/2)^2 = 0
ab + (A^2)/4 = 0
Therefore: ab = -(A^2)/4 = -1/4 * A^2
Sorry if it's hard to read. Rats i was late again... :P
^ This is the result if you have too much time on your hands + procrastination... ^
Hi, is anyone able to do question 17 d from exercise 10E ?
Thank you!
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If N is to the left of U, the answers must all be in the form ____N____U____, where the underscores represent places the remaining four letters can go. So:
Consider _N_U_, where we are trying to insert M. There are 3 places for it to go.
Now consider what will be left, we have: _N_U_M_, or _N_M_U_, or _M_N_U_. Regradless of where we place the M, there are four places for the next letter.
Following this, wherever we place the next letter will take away one available space, but create two either side of it, netting one more. So for M there are 3, for B there are 4, for E there are 5, and for R there are 6.
Doing it this way gives an answer of 3*4*5*6=360, and with this method we cover the event of placing the letters down in different orders.
http://prnt.sc/dfvluh pls halp. what does this ques mean
ATAR '17: 98
UNSW WAM: 86.5 :S
It's asking what values of a, b, and c are required to change the degree of the polynomial. For example, if you had a polynomial , and you wanted a polynomial of degree 1, a and b would both have to be zero, so that the polynomial became , with the degree of a polynomial being the highest exponent.
Can anyone help me question 23 of Exercise 1 E
Last edited by Drongoski; 26 Dec 2017 at 8:14 AM.
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