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In a box there are 4 ball point pens, and 3 fountain pens. How many 29 May 2018, 01:59
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

61% (02:35) correct 39% (02:41) wrong based on 463 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



To solve question 2: P&C Practice Question 2

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Article-1: Learn when to “Add” and “Multiply” in Permutation & Combination questions
Article-2: Fool-proof method to Differentiate between Permutation & Combination Questions
Article-3: 3 deadly mistakes you must avoid in Permutation & Combination

All Articles: Must Read Articles and Practice Questions to score Q51 !!!!
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B
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In a box there are 4 ball point pens, and 3 fountain pens. How many 29 May 2018, 02:55
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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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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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In a box there are 4 ball point pens, and 3 fountain pens. How many 29 May 2018, 04:25
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answer must be 44 as it is "or" question
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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.

Answer: B
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In a box there are 4 ball point pens, and 3 fountain pens. How many 16 Feb 2019, 10:14
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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
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In a box there are 4 ball point pens, and 3 fountain pens. How many 18 May 2020, 23:18
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why are we assuming that all the pens are different? It didnt mentioned so in the question.
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In a box there are 4 ball point pens, and 3 fountain pens. How many 30 May 2021, 06:47
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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
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In a box there are 4 ball point pens, and 3 fountain pens. How many 09 Oct 2025, 13:26
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