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Challenge and also a help..... Check image for question

Mathematics
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look at the 10th row in pascal's triangle : |dw:1438086192297:dw|
We want to find \(x\) such that \[\large ^{10}C_{x-1} - 3 *^{10}C_{x}~\gt~ 0\]
yes i m looking
|dw:1438086392992:dw|
yes >0 , understood that because sqare root function and denominator
notice that 45-3*10 > 0 10-3*1 > 0 1 - 3*0 > 0 so, x = 9, 10, 11 are fine
The answer is none of these... so I am confused
x = 8 doesn't work because 120 - 3*45 < 0 similarly x>11 doesn't work because 0 - 3*0 = 0
answer is \(a\) to my knowledge
let me analyse once, give me some time then I may ask you if you have no problem @ganeshie8
sure
http://www.wolframalpha.com/input/?i=solve+%2810+choose+%28x-1%29%29+-+3*%2810+choose+x%29+%3E+0+over+integers
Is this the language for combinations? Thats interesting.... hmmm...
solve (10 choose (x-1)) - 3*(10 choose x) > 0 over integers Can you explain this sentence with respect ot combination?
#to#
\(\large ^nC_r\) is read as \(\large n~~\text{choose}~~r\)
it is called binomial coefficient and represented by symbol : \[\large \binom{n}{r}\]
Each of the different groups or selections which can be made by taking some or all of a number of things(irrespective of order) //Can you please elaborate this sentence
// Its for Combination
45-3*10 > 0 You checked this for 9?
yes that is for x = 9
suppose you went for shopping and liked 5 shirts
ok
but you can only buy 3 of them
ok
how many ways are there to choose 3 shirts from the 5 shirts ?
for definiteness, lets call the shirts : \(a,~b,~c,~d,~e\)
calculations say 10
can you list them all ?
ok go on... nice
no, I don't have that strong concept.... I would definitely understand from you.... nice....
no, just list them all, its a simple counting problem
you can buy the shirts \(a,b,c\) or \(b,c,d\) or \(c,d,e\) or ... try listing them all
okay ... abc acd bed bcd ade bec cde aeb
more two
looks two are missing yeah
I am trying :P
bae
no no not bae
``` {a,b,c} {a,b,d} {a,c,d} {b,c,d} {a,b,e} {a,c,e} {b,c,e} {a,d,e} {b,d,e} {c,d,e} ```
\(\large ^{5}C_3\) gives the number of ways you can choose 3 shirts from 5
Thank you.... any method to find the terms in order... like you easily sound out the 10. suppose 16 or 24... any trick is there?
##found##
i hope you're familiar with the formula for computing \(^5C_3\) : \[^5C_3 = \dfrac{5!}{3!(5-3)!} =\dfrac{5!}{3!*2!}= \dfrac{5\times 4\times 3\times 2\times 1}{3\times 2\times 1*2\times 1} = 10\]
yes yes.... ofcourse
I am talking about counting.. thats hard! :P
`Thank you.... any method to find the terms in order... like you easily sound out the 10. suppose 16 or 24... any trick is there?` hey i didnt get what you're asking..
suppose abcdefgh and we have to choose 3 pairs then is there any way by counting? Like suppose I want to view all possibilities
you want see the entire list of 56 choices ? ``` {a,b,c} {a,b,d} {a,c,d} {b,c,d} {a,b,e} {a,c,e} {b,c,e} {a,d,e} {b,d,e} {c,d,e} {a,b,f} {a,c,f} {b,c,f} {a,d,f} {b,d,f} {c,d,f} {a,e,f} {b,e,f} {c,e,f} {d,e,f} {a,b,g} {a,c,g} {b,c,g} {a,d,g} {b,d,g} {c,d,g} {a,e,g} {b,e,g} {c,e,g} {d,e,g} {a,f,g} {b,f,g} {c,f,g} {d,f,g} {e,f,g} {a,b,h} {a,c,h} {b,c,h} {a,d,h} {b,d,h} {c,d,h} {a,e,h} {b,e,h} {c,e,h} {d,e,h} {a,f,h} {b,f,h} {c,f,h} {d,f,h} {e,f,h} {a,g,h} {b,g,h} {c,g,h} {d,g,h} {e,g,h} {f,g,h} ```
as you can see its not so pleasant to list them all... why would you want to do that lol
ofcourse it is unpleasant! but Maths is Maths.... but can you tell me how did you do it?
Can Wolfram do this?
https://jsfiddle.net/ganeshie8/r2hjr3js/1/
Thank you so much for giving your precious time.... :)
comb(n, set) lists all the combinations of size "n" from the set "set"
but trust me, you almost never want to list all the possibilities in math most of the time you only want the total number of choices, not the list.
ya its working.... thank you....
Ok got it! :P

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