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

Question

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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pushpitkc · Answer

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.

Chhavi28 · Answer

answer must be 44 as it is "or" question

EgmatQuantExpert · Answer

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

energetics · Answer

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

serenayong · Answer

why are we assuming that all the pens are different? It didnt mentioned so in the question.

Sanjaykumarro · Answer

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

bumpbot · Answer

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

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

Prep Toolkit

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