P&c q7

[Saumyojit]

A student is allowed to select at most n books from a collection of (2n+1) books. If the total number of ways in which he can select at least one book is 63, find the value of n.



Total no of ways he can select *Atleast* one book
=63
This line suggest that there are *six books all total*

Isn't it?

Each book will have two options of either selecting it or not selecting it.
So six books will have 2^6=64.
64-1=63.(atleast one)

So,2n+1=6
n=2.5?

N cannot be 2.5
That means there are 7 books of which 2 are alike .
So we consider two alike as only One book
Right?

At the time of selection we consider 6 books
So we get 2n+1=7

N=3

 

[Dr.Peterson]

A student is allowed to select at most n books from a collection of (2n+1) books. If the total number of ways in which he can select at least one book is 63, find the value of n.

The first step in problem solving is to figure out what it means. This one is not exactly easy to understand.

I take it to mean that 63 is the number of ways to choose some number from 1 through n out of 2n+1 books.

Total no of ways he can select *Atleast* one book
=63
This line suggest that there are *six books all total*

How does it "suggest" that?

As far as I can tell, you ignored the first sentence, and saw that 6 was the solution of the resulting (different) problem because it is the solution to 2^n - 1 = 63.

You can't bypass that step of interpreting the problem!

 

[Steven G]

In how many ways can he choose one book from 2n+1 books?
In how many ways can he choose two books from 2n+1 books?
In how many ways can he choose three books from 2n+1 books?
In how many ways can he choose four books from 2n+1 books?
In how many ways can he choose five books from 2n+1 books?
...
In how many ways can he choose n books from 2n+1 books?

Add these up and then .....?

 

[Saumyojit]

The first step in problem solving is to figure out what it means. This one is not exactly easy to understand.

I take it to mean that 63 is the number of ways to choose some number from 1 through n out of 2n+1 books.


How does it "suggest" that?

As far as I can tell, you ignored the first sentence, and saw that 6 was the solution of the resulting (different) problem because it is the solution to 2^n - 1 = 63.

You can't bypass that step of interpreting the problem!

One of my friend showed this process and I exactly copied

Sum nC0 to ncn = 2^n


(2n+1)c0 + (2n+1)c1 + (2n+1)c2 + (2n+1)c3.....(2n+1)Cn
= (2n+1)c(n+1) + (2n+1)c(n+2) + (2n+1)c(n+3) .....(2n+1)C(2n+1)




I don't understand
How they are equivalent?
If n=3 and 2n+1=7

First series gives 64 .

Next series sum gives something else

Or he is telling me something else.??

 

[pka]

One of my friend showed this process and I exactly copied
Sum nC0 to ncn = 2^n
I don't understand
How they are equivalent?
If n=3 and 2n+1=7
First series gives 64 .
Next series sum gives something else
Or he is telling me something else.??

You problem with all that you have posted is that you are in water way over your head and you don't know how to swim.
[MATH]{(a + b)^n} = \sum\limits_{k = 0}^n {\dbinom{n}{k}{x^k}{y^{n - k}}}[/MATH] is the binominal.
If [MATH]n=6,~x=1,~\&~y=1[/MATH] then we have [MATH]{(2)^6} = \sum\limits_{k = 0}^6 {\dbinom{6}{k}} [/MATH]That is the number of subsets of a set of six elements. BUT that number includes the empty-set.
In this case we want the non-empty sets so that [MATH]2^6-1=63[/MATH] non-empty subsets.

 

[Steven G]

One of my friend showed this process and I exactly copied

Sum nC0 to ncn = 2^n


(2n+1)c0 + (2n+1)c1 + (2n+1)c2 + (2n+1)c3.....(2n+1)Cn
= (2n+1)c(n+1) + (2n+1)c(n+2) + (2n+1)c(n+3) .....(2n+1)C(2n+1)




I don't understand
How they are equivalent?
If n=3 and 2n+1=7

First series gives 64 .

Next series sum gives something else

Or he is telling me something else.??

Sum nC0 to ncn = 2^n is correct. How is this helpful?
You want sum (2n+1)C1 to (2n+1)Cn

 

[Dr.Peterson]

One of my friend showed this process and I exactly copied

Sum nC0 to ncn = 2^n

(2n+1)c0 + (2n+1)c1 + (2n+1)c2 + (2n+1)c3.....(2n+1)Cn
= (2n+1)c(n+1) + (2n+1)c(n+2) + (2n+1)c(n+3) .....(2n+1)C(2n+1)

I don't understand
How they are equivalent?
If n=3 and 2n+1=7

First series gives 64 .

Next series sum gives something else

Or he is telling me something else.??

Are you aware of the fact that nCr = nC(n-r)? The LHS is the number of ways to choose r items of the n, while the RHS is the number of ways to choose the remaining n-r items -- so they are equivalent. In particular, each term of your LHS is equal to the opposite term in your RHS.

For your example, the LHS of your equation, for your example, is 7C0 + 7C1 + 7C2 + 7C3 = 1 + 7 + 21 + 35, while the RHS is 7C4 + 7C5 + 7C6 + 7C7 = 35 + 21 + 7 + 1. They are obviously equal, both 64. And the sum of both sides is the sum of one entire row of Pascal's Triangle, namely 2^7 = 128 (all ways to choose some of 7 items).

Did you really calculate both sides, or just assume they didn't look the same?

I will tell you that this fact is one thing I tried in order to find a solution to your problem, which involves the LHS; but I haven't pursued it all the way.

 

[Saumyojit]

Are you aware of the fact that nCr = nC(n-r)? The LHS is the number of ways to choose r items of the n, while the RHS is the number of ways to choose the remaining n-r items -- so they are equivalent. In particular, each term of your LHS is equal to the opposite term in your RHS.

For your example, the LHS of your equation, for your example, is 7C0 + 7C1 + 7C2 + 7C3 = 1 + 7 + 21 + 35, while the RHS is 7C4 + 7C5 + 7C6 + 7C7 = 35 + 21 + 7 + 1. They are obviously equal, both 64. And the sum of both sides is the sum of one entire row of Pascal's Triangle, namely 2^7 = 128 (all ways to choose some of 7 items).

Did you really calculate both sides, or just assume they didn't look the same?

I will tell you that this fact is one thing I tried in order to find a solution to your problem, which involves the LHS; but I haven't pursued it all the way.

Now I got it .
You used this concept NCR=NC(n-r)

We know that sum of 2n+1C0 +...+2n+1Cn=
64

Now we know that each term above is equal to their Corresponding (n-r)
Term
So that series Sum will give me *64* Also

So, 128 is the no of ways of selecting any no of objects from 2n+1 Terms.

So, 2^(2n+1)=128

Now half of this is 64 .

2^(2n+1)/2= 64

Atleast one, so subtract one from both sides

2^(2n+1) /2 -1=63

N=3.

Nice sum ?

 

[Dr.Peterson]

So, 2^(2n+1)=128

Now half of this is 64 .

2^(2n+1)/2= 64

Atleast one, so subtract one from both sides

2^(2n+1) /2 -1=63

N=3.

Your work doesn't show how you actually find the answer. I'd do it this way:

2^(2n+1) = 128 = 2^7​

2n+1 = 7​

2n = 6​

n = 3​