How many possible 7-digit codes can be formed from the lette

This topic has expert replies
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

[GMAT math practice question]

How many possible 7-digit codes can be formed from the letters s, u, c, c, e, s, s?

A. 400
B. 420
C. 450
D. 500
E. 510

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

by Brent@GMATPrepNow » Tue Apr 24, 2018 5:15 am
Max@Math Revolution wrote:[GMAT math practice question]

How many possible 7-digit codes can be formed from the letters s, u, c, c, e, s, s?

A. 400
B. 420
C. 450
D. 500
E. 510
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

-------NOW ONTO THE QUESTION--------------------

In the word SUCCESS,
There are 7 letters in total
There are 2 identical C's
There are 3 identical S's
So, the total number of possible arrangements = 7!/[(2!)(3!)]
= 420

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image

User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

by Max@Math Revolution » Thu Apr 26, 2018 12:12 am
=>

The number of permutations of 7 different letters is 7!.
However, 3 of the letters are 's' and 2 of the letters are 'c'. Therefore, the number of permutations of the letters s, u, c, c, e, s, s is 7! / (3! * 2!) = 420.

Therefore, the answer is B.
Answer: B