Problem Solving

jack0997 wrote:A botanist plans to code each experimental plant used in an experiment with a code that consists of either a single letter or a pair of distinct letters written in an alphabetic order. What is the least number of letters that can be used if there are 15 plants, and each plant is to get a different code?

(A) 3 (B) 4 (C) 5 (D) 7 (E) 14

Hi jack,

Let the number of letters to be used be n.

The number of plants that can be identified using a single letter = nC1 = n.

The number of plants that can be identified using two distinct letters = nC2 = [n.(n − 1)] / (1.2) (since the letters are to kept in alphabetic order, we must not order them or apply nP2.)

Thus, total number of plants that can be identified if we attempt to have all the 15 codes that are either one-letter code or two-letter codes = n + n (n − 1)/2

Since we need to have at least 15 identifications, we have:

n + n (n − 1)/2 ≥ 15

=> [2n+n(n−1)]/2 ≥ 15

=> n(2+(n−1)) ≥ 30

=> n(n+1) ≥ 30

Working with the options and starting with the least value of n, we see that [spoiler]n = 5 [/spoiler]satisfies the above inequality.

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide

-Jay
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Oops. I made a mistake. I hurriedly read the question and interpreted 'either a single letter or a pair of distinct letters' as 'either a single letter or a string of distinct letters.' Mitch cannot be wrong. Corrected my post.

-Jay
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Here's a similar (official) GMAT question to practice with: https://www.beatthegmat.com/distribution ... 63732.html

Cheers,
Brent

GMATGuruNY wrote:

Jay@ManhattanReview wrote:Mitch cannot be wrong.

I'm going to frame this post and mount it on the wall so that it is visible to my family at all times.

That's a nice decision you've taken. At least this way you can win an argument with your wife.

-Jay
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Manhattan Review GMAT Prep

Locations: New York | Prague | Johannesburg | Lisbon | and many more...

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