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

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

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New post 25 Jun 2017, 12:15
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6
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A
B
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D
E

Difficulty:

  55% (hard)

Question Stats:

59% (02:28) correct 41% (01:57) wrong based on 132 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
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Joined: 25 Feb 2013
Posts: 1210
Location: India
GPA: 3.82
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Re: John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

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New post 25 Jun 2017, 12:30
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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
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Re: John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

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New post 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)
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John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

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New post 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 \(= 1+5+10+10+5+1 = 32\)

Answer (D)...
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Re: John will exhibit 6 pictures including the work X. If he will exhibit  [#permalink]

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New post 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
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Re: John will exhibit 6 pictures including the work X. If he will exhibit   [#permalink] 08 Jan 2019, 13:01
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John will exhibit 6 pictures including the work X. If he will exhibit

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