At a certain laboratory, chemical substance are identified by an unordered combination of three different...

This topic has expert replies
Moderator
Posts: 2505
Joined: Sun Oct 15, 2017 1:50 pm
Followed by:6 members

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

Source: Veritas Prep

At a certain laboratory, chemical substance are identified by an unordered combination of three different colors. If no chemical may be assigned the same three colors as any other, what is the maximum number of substances that can be identified using seven colors?

A. 21
B. 35
C. 105
D. 135
E. 210

The OA is B
Source: — Problem Solving |

Legendary Member
Posts: 2214
Joined: Fri Mar 02, 2018 2:22 pm
Followed by:5 members
In permutation and combination, an unordered combination is of the form nCr, which has the formula =>
$$\frac{n!}{r!\cdot\left(n-r\right)!}$$
Total colors available = 7
And we need to choose 3 colors. So, we find 7C3
$$7C_3=\frac{7!}{3!\cdot\left(7-3\right)!}$$
$$=\frac{7!}{3!4!}$$
$$=\frac{7\cdot6\cdot5\cdot4!}{3\cdot2\cdot1\cdot4!}=\frac{7\cdot6\cdot5}{3\cdot2}=35$$
Answer = option B

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8086
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members
BTGmoderatorLU wrote:
Sat Dec 19, 2020 3:27 am
Source: Veritas Prep

At a certain laboratory, chemical substance are identified by an unordered combination of three different colors. If no chemical may be assigned the same three colors as any other, what is the maximum number of substances that can be identified using seven colors?

A. 21
B. 35
C. 105
D. 135
E. 210

The OA is B
Solution:

The number of ways of choosing 3 colors out of 7 (where order doesn’t matter) is 7C3 = 7! / (3! x 4!) = (7 x 6 x 5) / (3 x 2) = 7 x 5 = 35.

Answer: B

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage