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Bunuel
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Jared will pick 3 friends to join him on a road trip. His friend group 08 Jul 2019, 02:37
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C
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25% (medium)

Question Stats:

75% (01:53) correct 25% (02:22) wrong based on 167 sessions

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Jared will pick 3 friends to join him on a road trip. His friend group consists of 4 musicians and 3 poets. In how many different ways can Jared select his 3 traveling companions so that he has at least one musician and at least one poet among them?

(A) 16
(B) 18
(C) 30
(D) 36
(E) 84
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C
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Jared will pick 3 friends to join him on a road trip. His friend group 08 Jul 2019, 02:50
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Jared will pick 3 friends to join him on a road trip. His friend group consists of 4 musicians and 3 poets. In how many different ways can Jared select his 3 traveling companions so that he has at least one musician and at least one poet among them?

(A) 16
(B) 18
(C) 30
(D) 36
(E) 84
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Possible selections = 1M & 2P (or) 2M & 1P
= 4c1*3c2 + 4c2*3c1
= 4*3 + 6*3
= 30

IMO Option C

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Jared will pick 3 friends to join him on a road trip. His friend group 08 Jul 2019, 09:58
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Bunuel
Jared will pick 3 friends to join him on a road trip. His friend group consists of 4 musicians and 3 poets. In how many different ways can Jared select his 3 traveling companions so that he has at least one musician and at least one poet among them?

(A) 16
(B) 18
(C) 30
(D) 36
(E) 84
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total combination ; 4c1*3c2+4c2*3c1 ; 12+18; 30
IMO C
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Jared will pick 3 friends to join him on a road trip. His friend group 11 Jul 2019, 20:02
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Bunuel
Jared will pick 3 friends to join him on a road trip. His friend group consists of 4 musicians and 3 poets. In how many different ways can Jared select his 3 traveling companions so that he has at least one musician and at least one poet among them?

(A) 16
(B) 18
(C) 30
(D) 36
(E) 84
Show more

We have two scenarios:

Scenario 1: 2 poets and 1 musician

2 poets can be selected in 3C2 = 3 ways.

1 musician can be selected in 4C1 = 4 ways.

Therefore, in this scenario we have a total of 3 x 4 = 12 ways.

Scenario 2: 2 musicians and 1 poet

2 musicians can be selected in 4C2 = (4 x 3)/2! = 6 ways.

1 poet can be selected in 3C1 = 3 ways.

Therefore, in this scenario we have a total of 6 x 3 = 18 ways.

The number of ways to select the friends with at least one poet and one musician is 12 + 18 = 30 ways.

Answer: C
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Jared will pick 3 friends to join him on a road trip. His friend group 11 Jul 2019, 22:28
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Jared will pick 3 friends to join him on a road trip. His friend group consists of 4 musicians and 3 poets. In how many different ways can Jared select his 3 traveling companions so that he has at least one musician and at least one poet among them?

(A) 16
(B) 18
(C) 30
(D) 36
(E) 84
Show more

# of different ways of selecting travel companions = \(C^4_1 * C^3_2 + C^4_2 * C^3_1 = 4*3 + 6*3 = 30\)

IMO C
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Jared will pick 3 friends to join him on a road trip. His friend group 19 Jul 2019, 12:51
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What's the wrong with the following?

1 musician = 4C1
1 poet = 3C1
1 any other = 5C1

Total == 4C1*3C1*5C1 == 60

What am I screwing up?

Help would be appreciated.

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Jared will pick 3 friends to join him on a road trip. His friend group 17 Aug 2019, 02:14
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The above is a good question. Since 60 is two times 30, you must double counting somehow, but not sure how!
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Jared will pick 3 friends to join him on a road trip. His friend group 17 Aug 2019, 02:48
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7C3=35 total ways to select 3 people among 7 ones.
4C3=4, If all people are selected from 4 musicans.
3C3=1, if all people are selected
from 3 poets.
So, 35-(4+1)=30
Option C

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Jared will pick 3 friends to join him on a road trip. His friend group 11 Nov 2023, 12:17
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