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What is the total number of permutations of n different things taken n

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Bunuel
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What is the total number of permutations of n different things taken n [#permalink] New post   01 Mar 2021, 12:34
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What is the total number of permutations of n different things taken not more than p times, when each thing may be repeated any number of times?


A. \(n^p +1\)

B. \(n(n^p-1)\)

C. \(\frac{n^p (n+1)}{(n-1)}\)

D. \(\frac{n(n^p - 1)}{ (n-1)}\)

E. \(n^p (n+1)\)

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Re: What is the total number of permutations of n different things taken n [#permalink] New post   01 Mar 2021, 20:25
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Bunuel wrote:
What is the total number of permutations of n different things taken not more than p times, when each thing may be repeated any number of times?


A. \(n^p +1\)

B. \(n(n^p-1)\)

C. \(\frac{n^p (n+1)}{(n-1)}\)

D. \(\frac{n(n^p - 1)}{ (n-1)}\)

E. \(n^p (n+1)\)

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Let us analyse the question:
It tells us the following things
a) First, there are n different things
b) Second, we can take one item or 2 items or 3 items .....till P items
c) When we pick these items, they can be same
d) We are looking at Permutation, so order makes a difference.

Solution:

If we take 1 item, it can be filled in n ways.
If we take 2 items, it can be filled in n*n ways.
If we take 3 items, it can be filled in n*n*n ways.
.........
If we take p items, it can be filled in \(n*n*....p \ \ times=n^p\)ways.

Thus total ways = \(n+n^2+n^3+....n^p\)

This is a geometric progression where the common ratio by which each subsequent term increases is n.
Formula for sum of a GP=\(\frac{First\\term(ratio^{number \\of \\terms}-1)}{ratio-1}\)

Here we have n as the first term, n as the common ratio and p as the number of terms.
Permutations =Answer =\(\frac{n(n^p-1)}{n-1}\)

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testtakerstrategy
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Re: What is the total number of permutations of n different things taken n [#permalink] New post   29 Dec 2021, 14:59
Is it just me or is this question strangely worded? Is this wording for "... taken not more than p times," common? This threw me off.
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IanStewart
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Re: What is the total number of permutations of n different things taken n [#permalink] New post   29 Dec 2021, 15:38
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testtakerstrategy wrote:
Is it just me or is this question strangely worded? Is this wording for "... taken not more than p times," common? This threw me off.


You'll sometimes see wording like this in a textbook for a third or fourth year undergrad combinatorics course, if you take a Math degree. You'd never see a question worded this way before you reach that level, and absolutely never on the GMAT. I taught undergrad combinatorics (well, TA'ed it, so I taught problem solving sessions) and I find questions worded this way almost incomprehensible, so there's no need to worry about it! Not sure of the source, but it's definitely not one someone should use for GMAT prep.
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Re: What is the total number of permutations of n different things taken n [#permalink] New post   29 Dec 2021, 16:45
Asked: What is the total number of permutations of n different things taken not more than p times, when each thing may be repeated any number of times?

The total number of permutations of n different things taken not more than p times = \(n + n^2 + .... + n^p = n(n^p-1)/(n-1)\)

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Re: What is the total number of permutations of n different things taken n [#permalink] New post   30 Dec 2021, 10:01
Pick some easy numbers.

Let's pick 4 things, taken at most 2 at a time.

Since order matters and with replacement, there are

4*4= 16 ways to pick 2 items.

Picking 1 at a time, there are 4 ways.

Total ways equals 20.


Testing the answers, answer D provides:

4(4^2-1)/(4-1) = 4*15/3 = 20, matching the answer above.

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Re: What is the total number of permutations of n different things taken n [#permalink]
30 Dec 2021, 10:01
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