GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Jun 2019, 22:27

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# John will exhibit 6 pictures including the work X. If he will exhibit

Author Message
TAGS:

### Hide Tags

Intern
Joined: 28 Dec 2014
Posts: 9
GMAT 1: 710 Q47 V41
John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

### Show Tags

25 Jun 2017, 12:15
1
6
00:00

Difficulty:

55% (hard)

Question Stats:

59% (02:28) correct 41% (01:57) wrong based on 132 sessions

### HideShow timer Statistics

John will exhibit 6 pictures including the work X. If he will exhibit one or more pictures and must exhibit the work X, how many such ways are possible?

(A) 26
(B) 28
(C) 30
(D) 32
(E) 34
Retired Moderator
Joined: 25 Feb 2013
Posts: 1210
Location: India
GPA: 3.82
Re: John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

### Show Tags

25 Jun 2017, 12:30
1
1
aazt wrote:
John will exhibit 6 pictures including the work X. If he will exhibit one or more pictures and must exhibit the work X, how many such ways are possible?

(A) 26
(B) 28
(C) 30
(D) 32
(E) 34

As Jhon can exhibit one OR more picture (including X) so number of ways could be -

Exhibiting 1 picture "X" OR exhibiting 2 Pictures OR exhibiting 3 pictures OR exhibiting 4 pictures OR exhibiting 5 pictures OR exhibiting 6 pictures

= 1 + 5C1 + 5C2+ 5C3 + 5C4+ 5C5 (as "X" is fixed, we are left with 5 pictures to choose from)
= 1+5+10+10+5+1 = 32

Option D
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3385
Location: India
GPA: 3.12
Re: John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

### Show Tags

25 Jun 2017, 12:39
1
Since John can exhibit 1 or more of his 6 pictures, it can be done in

1(if all the pictures are displayed)
1(if only X is displayed)
1*5c1 = 5 (if X and 1 of the other 5 are displayed)
1*5c2 = 10 (if X and 2 of the other 5 are displayed)
1*5c3 = 10 (if X and 3 of the other 5 are displayed)
1*5c4 = 5(if X and 4 of the other 5 are displayed)

Total ways : 1+5+10+10+5+1 = 32(Option D)
_________________
You've got what it takes, but it will take everything you've got
Director
Joined: 04 Dec 2015
Posts: 742
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)
John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

### Show Tags

01 Jul 2017, 01:04
1
aazt wrote:
John will exhibit 6 pictures including the work X. If he will exhibit one or more pictures and must exhibit the work X, how many such ways are possible?

(A) 26
(B) 28
(C) 30
(D) 32
(E) 34

John has 6 pictures.

Exhibiting 6 pictures could be in following ways;

He must exhibit picture X. Possible ways for $$X = 1$$

$$5_C_1 = \frac{5!}{4!1!} = 5$$

$$5_C_2 = \frac{5!}{3!2!} = 10$$

$$5_C_3 = \frac{5!}{2!3!} = 10$$

$$5_C_4 = \frac{5!}{4!1!} = 5$$

$$5_C_5 = \frac{5!}{5!} = 1$$

Therefore total number of ways John can exhibit the pictures $$= 1+5+10+10+5+1 = 32$$

Intern
Joined: 19 Aug 2018
Posts: 2
Re: John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

### Show Tags

08 Jan 2019, 13:01
Total ways of exhibiting pictures= 2^6 (2 options for each picture=yes/no) :64
Total ways when picture X is not exhibited= 1*2^5 : 32
So total ways when X is always included= 64-32: 32
Re: John will exhibit 6 pictures including the work X. If he will exhibit   [#permalink] 08 Jan 2019, 13:01
Display posts from previous: Sort by