The tribonacci sequence is a sequence of numbers where each term after

Question

The tribonacci sequence is a sequence of numbers where each term after the third is the sum of the three preceding terms. List T includes the first 15 numbers of the tribonacci sequence, starting with \(t_1 = 0\), \(t_2 = 0\), and \(t_3 = 1\). How many ways can three terms be selected from List T such that their sum is odd?

A. 28
B. 35
C. 196
D. 224
E. 231

A. 28
B. 35
C. 196
D. 224
E. 231

This is a  PS Butler Question

Check the links to other Butler Projects: 
 
   Data Sufficiency Butler 
   Critical Reasoning Butler  
   Sentence Correction Butler  
   Reading Comprehension Butler  
   Integrated Reasoning Butler

gmatophobia · Answer

Let's write the first few terms of the series

t_1  =  even 
 t_2  =  even 
 t_3  =  odd 
 t_4  = even + even + odd =  odd

t_5  = odd + odd + even =  even 
 t_6  = even + odd + odd =  even 
 t_7  = even + even + odd =  odd 
 t_8  = odd + even + even =  odd

Inference:  → Each block of four consecutive numbers in the series has 2 even terms and 2 odd terms.

Hence, in the first 15 terms we have = 
 
 8 even terms  
 7 odd terms

Question:  → We are asked to select three terms such that the sum is odd

Case 1 : Select 2 even terms from 8 available even terms AND Select 1 odd term

The number of ways 2 terms can be selected from 8 available terms =  ^8C_2   ways 
The number of ways 1 terms can be selected from 1 available terms =  ^7C_1   ways

Total =  \frac{8*7}{2} * 7  = 196

Case 2 : Select 3 odd terms

The number of ways 3 terms can be selected from 7 available terms =  ^7C_3   ways

Total =  \frac{7 * 6 * 5 }{ 3 * 2 * 1}  = 35

Total Number of ways = 196 + 35 = 231

Option E

Oppenheimer1945 · Answer

Total even terms =8
Total odd terms=7
No of ways= odd odd odd or even even odd
= 7C3+8C2.7C1= 33.7=231

Posted from my mobile device

Kingrajan · Answer

The series would however have 2 terms as 1. So I think the answer should be 203. When we choose two even and one odd number, the number 1 can be chosen only once. So it should be 7C3 + 8C2*6 (not 7)

35+168=203

Series would be

0,0,1,1,2,4,7,13,24,44,81,149,274,504,927

Posted from my mobile device

samarpan.g28 · Answer

This is the series: 0,1,2,3,6,11,20,37,68,125,230,423,778,1431,2632 ->15 terms.
Here, we have 7 odd and 8 even terms (0 is even).
The question asks in how many ways if we pick three numbers from the series, the sum of the three numbers will be an odd number.
There are 2 ways we can do it:
a. pick 3 odd numbers.
b. pick two even numbers and one odd number.

For a, 7C3=35.
For b, 8C2 * 7C1=196.

Therefore, a total of 35+196=231 ways.

I hope this helps.

The tribonacci sequence is a sequence of numbers where each term after

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

The tribonacci sequence is a sequence of numbers where each term after

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards