A botanist plans to code each experimental plant used in an experiment

Question

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) 15

(A) 3
(B) 4
(C) 5
(D) 7
(E) 15

Palak74 · Answer

I think the answer is option C

A - A 
B - AB & B 
C - AC, BC, C 
D - AD, BD, CD, D
E - AE,BE,CE,DE,E

Therefore 5 Letters needed

stne · Answer

If we have  3  letters then:

Single letter codes that can be formed :  3 
Double letter codes that can be formed:  3_C_2 
Total =  3+3 =6   codes

Total  6  codes can be formed = NOT SUFF. to code  15  plants

If we have  4  letters then:

Single letter codes : 4 
Double letter codes : 4_C_2

Total codes :  4+6 =10  , NOT SUFF. to code  15  plants

If we have   5  letters then:

Single letter codes  :5 
Double letter codes : 5_C_2=10

Total  5+10 =15  codes , Can be used to code the  15  plants.

Hence we need a min. of  5  letters to code  15  plants.

Ans-C

Hope it's clear.

RastogiSarthak99 · Answer

I guess a similar question is this: https://gmatclub.com/forum/a-company-th ... 68427.html Per the question, I'm arriving at n + nc2 >= 15 for only n=5 solves the least min required.

Edskore · Answer

This is a smart combinatorics problem that tests whether you can count systematically without overcounting or undercounting.

Here's the step-by-step approach:

Step 1: Understand what codes are available 
With n letters, you can create:
- Single-letter codes: n codes (A, B, C, D, E...)
- Two-letter codes: C(n,2) = n(n-1)/2 codes (AB, AC, AD, BC, BD...)

The two-letter codes use  combinations , not permutations, because the problem specifies "alphabetic order" — so AB is valid but BA isn't a separate code.

Step 2: Set up the inequality 
We need: n + n(n-1)/2 ≥ 15

Simplifying: n + n2/2 - n/2 ≥ 15
Multiply by 2: 2n + n2 - n ≥ 30
Simplify: n2 + n ≥ 30

Step 3: Test the answer choices 
- n = 3: 32 + 3 = 12 (not enough)
- n = 4: 42 + 4 = 20 (not enough)
- n = 5: 52 + 5 = 30 (exactly 15 codes!)

With 5 letters, you get:
• 5 single-letter codes: A, B, C, D, E
• 10 two-letter codes: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE
• Total: 15 codes ✓

Common trap:  Students sometimes count permutations (AB and BA as separate), which would give n2 total instead of n(n-1)/2. But the phrase "alphabetic order" is your clue that order matters — we only count each pair once.

Takeaway:  When you see "alphabetic order" or "distinct pairs," think combinations (C(n,2)), not permutations (P(n,2)).

The answer is  C. 5 .

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