Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A cube has eight vertices. How many different combinations are there where you can choose three of these vertices so that none of them are on the same cube face?  <div

A cube has eight vertices. How many different combinations are there where you can choose three of these vertices so that none of them are on the same cube face?

 

Option: 1

24


Option: 2

32


Option: 3

64


Option: 4

48


Answers (1)

best_answer

Given that,

There are 8 vertices in a cube.

Three vertices from the cube can be selected in a total of \mathrm{ { }^8 C_3 } different ways. 

Thus,

\mathrm{ { }^8 C_3=\frac{8 !}{3 ! 5 !} }

\mathrm{ { }^8 C_3=\frac{8 \times 7 \times 6}{3 \times 2} }

\mathrm{ { }^8 C_3=56 }

If the cube's three vertices are on the same face, there are six possible faces and \mathrm{{ }^4 C_3=4} possible vertices, for a total of six times four or 24 possibilities.
As a result, there are \mathrm{56-24=32} alternative ways to choose three vertices as long as they are not all on the same face of the cube.

Therefore, the number of ways to choose three of the vertices so that none of them are on the same cube face is 32 ways.

Posted by

rishi.raj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Marks vs Percentile 2025: How many marks do you need to score 95+ percentile?
anam-khan

JEE Main 2025 session 1 official answer key out; challenge by February 6; how to calculate NTA score?
anam-khan

JEE Main 2025: NIT Raipur's CSE cut-offs drop; closing ranks in popular branches

Latest Question