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# In a box there are 4 ball point pens, and 3 fountain pens. How many

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2722
In a box there are 4 ball point pens, and 3 fountain pens. How many  [#permalink]

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29 May 2018, 01:59
1
1
4
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Difficulty:

55% (hard)

Question Stats:

66% (02:29) correct 34% (02:50) wrong based on 53 sessions

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In a box there are 4 ball point pens, and 3 fountain pens. How many possible selections can be formed which have at least 2 items of every types of pen?

A. 26
B. 44
C. 52
D. 78
E. 130

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In a box there are 4 ball point pens, and 3 fountain pens. How many  [#permalink]

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29 May 2018, 02:55
1
EgmatQuantExpert wrote:
In a box there are 4 ball point pens, and 3 fountain pens. How many possible selections can be formed which have at least 2 items of every types of pen?

A. 26
B. 44
C. 52
D. 78
E. 130

When we need a minimum of 2 pens of each type, these are the following options available:

When we have 2 ball-point pens
1. Along with 2 fountain pens - $$C_2^4 * C_2^3 = 6*3 = 18$$
2. Along with 3 fountain pens - $$C_2^4 * C_3^3 = 6*1 = 6$$

When we have 2 fountain pens
1. Along with 3 ball-point pens - $$C_2^3 * C_3^4 = 3*4 = 12$$
2. Along with 4 ball-point pens - $$C_2^3 * C_4^4 = 3*1 = 3$$

When we have 3 ball-point pens along with 3 fountain pens - $$C_3^3 * C_3^4 = 1*4 = 4$$

When we have all 7 pens - $$C_3^3 * C_4^4 = 1*1 = 1$$

The total possibilities are 18 + 6 + 12 + 3 + 4 + 1 = 44

Therefore, there are 44(Option B) possible selections which have at least 2 items of every types of pen.
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Joined: 21 Apr 2018
Posts: 2
Re: In a box there are 4 ball point pens, and 3 fountain pens. How many  [#permalink]

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29 May 2018, 04:25
answer must be 44 as it is "or" question
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2722
Re: In a box there are 4 ball point pens, and 3 fountain pens. How many  [#permalink]

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31 May 2018, 02:03

Solution

Given:
• In the box,
o Ball point pens = 4
o Fountain pens = 3

To find:
• How many possible selections can be formed which have at least 2 items of every types of pen

Approach and Working:
As we have to ensure at least 2 items of every types of pen, the different possible cases and their respective possibilities are as follows:
• 2 ball point pens and 2 fountain pens = $$^4C_2 * ^3C_2 = 6 * 3 = 18$$
• 3 ball point pens and 2 fountain pens = $$^4C_3 * ^3C_2 = 4 * 3 = 12$$
• 4 ball point pens and 2 fountain pens = $$^4C_4 * ^3C_2 = 1 * 3 = 3$$
• 2 ball point pens and 3 fountain pens = $$^4C_2 * ^3C_3 = 6 * 1 = 6$$
• 3 ball point pens and 3 fountain pens = $$^4C_3 * ^3C_3 = 4 * 1 = 4$$
• 4 ball point pens and 3 fountain pens = $$^4C_4 * ^3C_3 = 1 * 1 = 1$$

Therefore, total number of possible selections = 18 + 12 + 3 + 6 + 4 + 1 = 44

Hence, the correct answer is option B.

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Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
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Joined: 05 Feb 2018
Posts: 141
Re: In a box there are 4 ball point pens, and 3 fountain pens. How many  [#permalink]

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16 Feb 2019, 10:14
The question is asking for the #ways to choose at least 2 b and f pens

For b: For f:
4C0=1, 3C0=1
4C1=4, 3C1=3

4C2=6, 3C2=3
4C3=4, 3C1=1
4C4=1
+ =11,+ = 4

11 ways to chose b*4 ways to choose f = 44
Re: In a box there are 4 ball point pens, and 3 fountain pens. How many   [#permalink] 16 Feb 2019, 10:14
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