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

- arindameducationusc

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

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- arindameducationusc

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- arindameducationusc

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- arindameducationusc

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## More answers

- ganeshie8

look at the 10th row in pascal's triangle :
|dw:1438086192297:dw|

- ganeshie8

We want to find \(x\) such that
\[\large ^{10}C_{x-1} - 3 *^{10}C_{x}~\gt~ 0\]

- arindameducationusc

yes i m looking

- ganeshie8

|dw:1438086392992:dw|

- arindameducationusc

yes >0 , understood that because sqare root function and denominator

- ganeshie8

notice that
45-3*10 > 0
10-3*1 > 0
1 - 3*0 > 0
so, x = 9, 10, 11 are fine

- arindameducationusc

The answer is none of these... so I am confused

- ganeshie8

x = 8 doesn't work because
120 - 3*45 < 0
similarly x>11 doesn't work because
0 - 3*0 = 0

- ganeshie8

answer is \(a\) to my knowledge

- arindameducationusc

let me analyse once, give me some time then I may ask you if you have no problem @ganeshie8

- ganeshie8

sure

- ganeshie8

http://www.wolframalpha.com/input/?i=solve+%2810+choose+%28x-1%29%29+-+3*%2810+choose+x%29+%3E+0+over+integers

- arindameducationusc

Is this the language for combinations? Thats interesting.... hmmm...

- arindameducationusc

solve (10 choose (x-1)) - 3*(10 choose x) > 0 over integers
Can you explain this sentence with respect ot combination?

- arindameducationusc

#to#

- ganeshie8

\(\large ^nC_r\) is read as \(\large n~~\text{choose}~~r\)

- ganeshie8

it is called binomial coefficient and represented by symbol :
\[\large \binom{n}{r}\]

- arindameducationusc

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

- arindameducationusc

// Its for Combination

- arindameducationusc

45-3*10 > 0
You checked this for 9?

- ganeshie8

yes that is for x = 9

- ganeshie8

suppose you went for shopping and liked 5 shirts

- arindameducationusc

ok

- ganeshie8

but you can only buy 3 of them

- arindameducationusc

ok

- ganeshie8

how many ways are there to choose 3 shirts from the 5 shirts ?

- ganeshie8

for definiteness, lets call the shirts : \(a,~b,~c,~d,~e\)

- arindameducationusc

calculations say 10

- ganeshie8

can you list them all ?

- arindameducationusc

ok go on... nice

- arindameducationusc

no, I don't have that strong concept.... I would definitely understand from you.... nice....

- ganeshie8

no, just list them all, its a simple counting problem

- ganeshie8

you can buy the shirts \(a,b,c\) or \(b,c,d\) or \(c,d,e\) or ...
try listing them all

- arindameducationusc

okay ...
abc acd bed
bcd ade bec
cde aeb

- arindameducationusc

more two

- ganeshie8

looks two are missing yeah

- arindameducationusc

I am trying :P

- arindameducationusc

bae

- arindameducationusc

no no not bae

- ganeshie8

```
{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}
```

- ganeshie8

\(\large ^{5}C_3\) gives the number of ways you can choose 3 shirts from 5

- arindameducationusc

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?

- arindameducationusc

##found##

- ganeshie8

i hope you're familiar with the formula for computing \(^5C_3\) :
\[^5C_3 = \dfrac{5!}{3!(5-3)!} =\dfrac{5!}{3!*2!}= \dfrac{5\times 4\times 3\times 2\times 1}{3\times 2\times 1*2\times 1} = 10\]

- arindameducationusc

yes yes.... ofcourse

- arindameducationusc

I am talking about counting.. thats hard! :P

- ganeshie8

`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..

- arindameducationusc

suppose abcdefgh and we have to choose 3 pairs
then is there any way by counting? Like suppose I want to view all possibilities

- ganeshie8

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}
```

- ganeshie8

as you can see its not so pleasant to list them all... why would you want to do that lol

- arindameducationusc

ofcourse it is unpleasant! but Maths is Maths.... but can you tell me how did you do it?

- arindameducationusc

Can Wolfram do this?

- ganeshie8

https://jsfiddle.net/ganeshie8/r2hjr3js/1/

- arindameducationusc

Thank you so much for giving your precious time.... :)

- ganeshie8

comb(n, set)
lists all the combinations of size "n" from the set "set"

- ganeshie8

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.

- arindameducationusc

ya its working.... thank you....

- arindameducationusc

Ok got it! :P

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