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

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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]

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25 Jun 2017, 11:15
1
6
00:00

Difficulty:

65% (hard)

Question Stats:

61% (01:48) correct 39% (01:37) wrong based on 120 sessions

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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: 1220
Location: India
GPA: 3.82
Re: John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

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25 Jun 2017, 11:30
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: 3334
Location: India
GPA: 3.12
Re: John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

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25 Jun 2017, 11:39
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)
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Joined: 04 Dec 2015
Posts: 738
Location: India
Concentration: Technology, Strategy
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WE: Information Technology (Consulting)
John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

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01 Jul 2017, 00:04
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]

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08 Jan 2019, 12:01
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 &nbs [#permalink] 08 Jan 2019, 12:01
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