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- mathmath333

Counting question

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- mathmath333

\(\large \color{black}{\begin{align}
& \normalsize \text{A squad consists of 16 cricketers including Sachin and Rahul. }\hspace{.33em}\\~\\
& \normalsize \text{ In how many ways can the team of 11 cricketers be selected such that,}\hspace{.33em}\\~\\
& \normalsize \text{If Sachin is selected then Rahul is dropped and }\hspace{.33em}\\~\\
& \normalsize \text{if Rahul is selected then Sachin is dropped .}\hspace{.33em}\\~\\
\end{align}}\)

- mathmath333

- amilapsn

There are three possible kinds of situations can you guess them:

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## More answers

- mathmath333

there are 2 possiblities

- amilapsn

Question can be interpreted in many ways... ok.. tell me what your two possibilities are...

- mathmath333

Is S is there R is not vice versa

- mathstudent55

Since you can only make a team without one or without the other or without both, think of making a team of 11 out of 15 (no Sachin) and a team of 11 out of 15 (no Rahul).

- amilapsn

then what about both of them are not in the team?

- ganeshie8

isn't the answer simply \[\binom{16}{11} - \binom{14}{11}\]

- mathstudent55

\(2(_{15}C_{11})\)

- amilapsn

@ganeshie8 yes indeed that's another way of doing it...

- amilapsn

It's much simpler than my method...:)

- amilapsn

@mathmath333 did you get ganeshie's one?

- mathmath333

u mean this one
\(\Large \dbinom{16}{11} - \dbinom{14}{9}\)

- mathmath333

not really

- amilapsn

yep

- ganeshie8

it is 14 choose 11
not 14 choose 9

- mathmath333

but in book it is given this \(\Large \dbinom{16}{11} - \dbinom{14}{9}\)

- mathmath333

any way i didn't get how that comes

- amilapsn

it is \(\color{red}{16}\) choose 11
not 14 choose 9

- mathmath333

ok but how did u get this answer
\(\dbinom{16}{11} - \dbinom{14}{11}\)

- amilapsn

\( \dbinom{16}{11}\) means number of all teams that can be made without considering the condition.. ok?

- mathmath333

yes

- amilapsn

Then we just have to remove the number of teams where there are S and R both , right?

- ganeshie8

Ahh right, @amilapsn thanks for catching :)

- mathmath333

ok

- amilapsn

Then we are left with:
1.teams where there is not S or R
2.teams with only S
3 teams with only R
Condition fulfilled...

- mathmath333

Then we just have to remove the number of teams where there are S and R both
this will be done by \(-\dbinom{14}{11}\) ?

- amilapsn

Yes. Because We've already chosen dravid and thendulkar so we only have to choose anothe nine players out of the rest(14).. Got that?

- mathmath333

and by other method can u tell me
1.teams where there is not S or R
2.teams with only S
3 teams with only R

- amilapsn

1-we've choose 11 out of 14
2-we've already chosen S so we have to make sure that R is no there so now we have to choose 10 out of 12
3-the same as 2 we've to choose 10 out of 12

- mathmath333

2-we've already chosen S so we have to make sure that R is no there so now we have to choose 10 out of \(\color{red}{12}\)
3-the same as 2 we've to choose 10 out of \(\color{red}{12}\)
how does \(\color{red}{12}\) comes

- amilapsn

Let's consider the second situation... After choosing S how much players we are left with?

- mathmath333

15

- amilapsn

then can we get R to the team?

- mathmath333

no

- amilapsn

Then how much we've to consider?

- mathmath333

14

- amilapsn

So we've to choose 10 out of 14... (omg my bad that 12 should be 14... sorry for that mistake)

- mathmath333

then by the same logic wolfram gives this false
http://www.wolframalpha.com/input/?i=is+%5Cbinom%7B16%7D%7B11%7D-%5Cbinom%7B14%7D%7B11%7D%3D%5Cbinom%7B14%7D%7B10%7D%2B%5Cbinom%7B14%7D%7B10%7D%2B%5Cbinom%7B14%7D%7B11%7D+%3F

- mathmath333

while this is given true
http://www.wolframalpha.com/input/?i=is+%5Cbinom%7B16%7D%7B11%7D-%5Cbinom%7B14%7D%7B9%7D%3D%5Cbinom%7B14%7D%7B10%7D%2B%5Cbinom%7B14%7D%7B10%7D%2B%5Cbinom%7B14%7D%7B11%7D+%3F

- amilapsn

wait a sec... Let me see...

- amilapsn

Ok so what's the problem? We are alright, know?

- mathmath333

\(\dbinom{16}{11}-\dbinom{14}{11}\)
u said earlier that this is correct but wolfram isn't matching it with the other way

- amilapsn

that second 11 should be nine:
I've said that:\[\text{Yes. Because We've already chosen dravid and thendulkar so we only have to choose}\]
\[\text{anothe nine players out of the rest(14).. Got that?}\]

- mathmath333

ok lol i missed that comment earlier completely thanks

- amilapsn

Yw!

- ganeshie8

hows the test between india and lanka going

- amilapsn

We won the 1st one you know... It was a legendary match due to Chandimal's playing.. Today match is also a very important for all Sri Lankans because it is the final match of Sanga. We'll miss him so much...

- mathmath333

In the question why we didn't include the case both sachin and rahul are selected.

- amilapsn

Because the it's said in the question:\[\large \color{black}{\begin{align}
& \normalsize \text{A squad consists of 16 cricketers including Sachin and Rahul. }\hspace{.33em}\\~\\
& \normalsize \text{ In how many ways can the team of 11 cricketers be selected such that,}\hspace{.33em}\\~\\
& \normalsize \text{If Sachin is selected then Rahul is dropped and }\hspace{.33em}\\~\\
& \normalsize \text{if Rahul is selected then Sachin is dropped .}\hspace{.33em}\\~\\
\end{align}}\]

- mathmath333

by the way india lost 1st test against lanka

- mathmath333

are u from lanka

- amilapsn

yep.

- mathmath333

but it was also not stated that both sachin and rahul won't be selected, then also we add it

- amilapsn

But it doesn't break the condition right? We have to account for all situations without breaking the condition..

- mathmath333

but y it dont break the condition

- dan815

TotalWays - Ways of sachin and Rahul

- dan815

when you to 16 Choose 11
right now in all those combinations it includes the one where both of them are selected that is the only part we have to subtract

- amilapsn

That's the nature of "if".
Have you done logic @mathmath333 ?

- mathmath333

no

- dan815

so 16 Choose 11 - 14 Choose 9

- dan815

there are a couple other ways of adding the different cases, this is one way

- dan815

another way is to add
the cases of no sachin and rahul
then only sachin
then only rahul

- amilapsn

\[\href{
https://en.wikipedia.org/wiki/Material_conditional{\huge p\Rightarrow q}\]

- mathmath333

ok thanks i will look at this logic thing and interpretation of "if"

- amilapsn

It's pretty simple you'll get it easily.

- dan815

2*(14 Choose10) + 14 choose 11
^ ^________no rahul and sachil
'------ only rahul + only sachin also should work

- dan815

you can actually go on to show this actually always true
2*(14 Choose10) + 14 choose 11 = 16 choose 11 - 14 choose 9
2*(n choose k) + n choose (k+1) = (n+2) choose (k+1) - n choose(k-1)

- dan815

if u like that sorta thing lol.. try proving that

- dan815

|dw:1440177672216:dw|

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