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Question

A student is allowed to select at most n books from a collection of 2n+1 books. If the number of ways in which he can do this, is 64, then the value of n is equal to


A

6

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B

4

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C

3

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D

None of these

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Solution

The correct option is C

3


Explanation for the correct option :

Step-1 : (Find the total number of ways of choosing at most nbooks from a collection of 2n+1books)

Formula to be used : We know that the number ways of choosing r objects from a collection of n objects is Crn.

Given that, a student is allowed to select at most nbooks from a collection of 2n+1books. So, he can choose 0 book or 1book or 2 books or …. or at most nbooks.

Then, the number of ways he can choose 0 book is C02n+1.

The number of ways in which he can choose 1 book is C12n+1.

The number of ways in which he can choose 2 books is C22n+1.

The number of ways in which he can choose n books is Cn2n+1.

So, the total number of ways in which he can choose at most n books from 2n+1 books =C02n+1+C12n+1+C22n+1++Cn2n+1.

Step-2 : Evaluate the expression C02n+1+C12n+1++C2n+12n+1.

Formula to be used : We know that

Crn=Cn-rn and

C0n+C1n+C2n++Cnn=2n.

So, using the above formula, we get the following identities :

C02n+1=C2n+12n+1C12n+1=C2n2n+1C22n+1=C2n-12n+1Cn2n+1=Cn+12n+1

Then using the above identities, we obtain

C02n+1+C12n+1+C22n+1++Cn2n+1+Cn+12n+1++C2n-12n+1+C2n2n+1+C2n+12n+1=C02n+1+C12n+1+C22n+1++Cn2n+1+Cn2n+1++C22n+1+C12n+1+C02n+1=2C02n+1+C12n+1+C22n+1++Cn2n+1

Again, using the formula, we also have, C02n+1+C12n+1+C22n+1++Cn2n+1+Cn+12n+1++C2n-12n+1+C2n2n+1+C2n+12n+1=22n+1

Thus we must get the following identity :

2C02n+1+C12n+1+C22n+1++Cn2n+1=22n+1C02n+1+C12n+1+C22n+1++Cn2n+1=22n

Step-3 : (Evaluate the value of n)

Now, by the question

C02n+1+C12n+1+C22n+1++Cn2n+1=64.

and from Step-2, we get

C02n+1+C12n+1+C22n+1++Cn2n+1=22n.

Thus,

22n=6422n=262n=6n=3

Therefore, n=3.

Hence option (C) is the correct answer.


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