A shopkeeper keeps pens of two different companies Company A and Com

Question

A shopkeeper keeps pens of two different companies – Company A and Company B. Each company produces distinct pens and in 4 different colours: Red, Green, Blue, and Black and the shopkeeper keeps all these colours of both companies. James came to this shop to buy 2 pens, in how many ways can he buy pens of different colour?

A. 6
B. 12
C. 24
D. 40
E. 48

A. 6
B. 12
C. 24
D. 40
E. 48

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ManifestDreamMBA · Answer

There are a total of 8 options to pick the first pen
For second pen, James can pick any pen from the 7 remaining except the same color as picked in first. This gives him 6 options

Ways can he buy pens of different colour = 8*6/2 = 24 (Dividing by 2 to account for combinations)

Dereno · Answer

James need to buy 2 pens out of the total 8 pens , which are divided into 4 colours ( red, green, blue, and black) among two companies A and B each having the same colours.

First let’s select the 2 colours out of 4 colours 4C2 = 6
Then, for the selected colours say for example, red and green. For each colour we have two options company A or B. Two options each for red and green = 2*2=4

6*4 = 24  option C

Alter :

we have  2  options to select from company A or B. 
After selecting company, we have  4  options to choose our first colour . 
 ignoring the selected colour we are left with 3 colours.
 Hence, 2*4*3 = 24. Option C

ajay0520 · Answer

answer c=24 
total no of pen= 8
and total unique color = 4
he want to choose 2 different pen
 so for first pen he have 8 different way of choosing 
and for second pen he has 8-2=6 different way of choosing. as he wants both of different color

hence total possible no of selection  = 8*6/2
 = 24  (here we divide by 2 because its a combination question and the order in which he choose any pen doesnt matters)

GMATinsight · Answer

Varieties of Company A = 4 (all 4 colors)
Varieties of Company B = 4 (all 4 colors)
Total Different pens = 8

Total Choices for 2 colors = 4C2 = 6
Total Choices to choose pen from each color = 2(from 1st color)*2(from 2nd color) = 4

Total Favorable choices = 6*4 = 24

Answer: Option C

Related Video:

https://www.youtube.com/watch?v=qbN8UsktG5M

TC97 · Answer

The answer is 24. Two ways to solve this problem: GMAT-Club-Forum-f0zdd3xe.png

stne · Answer

In this question answer depends on how we interpret the question.

e.g. If R,G and G,R are two different ways to choose color, then answer will be   48

4_{C_2}*2 *2 *2 = 48

4_{C_2} * 2 =   Ways to choose  2  different pens and multiplying by  2 , to account for order.

2= There are two companies for the first pen chosen, thus two options for the first pen.

2= There are two companies for the second pen chosen, thus two options for the second pen.

Now if order does not matter e.g R,G and G,R are same then.

4_{C_2} * 2 *2 = 24

4_{C_2}=   Ways to choose two different pens, when order does not matter.

2= There are two companies for the first pen chosen

2= There are two companies for the second pen chosen

Hope it helps.

stne · Answer

If I may just point out.

In your alternate solution, although you have arrived at the same answer as your main solution.Both solutions do not use the same logic.

In your main solution, you have considered order not to be important. Well as in your alternate solution you have considered order to be important.

Alternate solution should ideally be  \frac{4*3}{2} *2 *2 = 24

4*3 =  Ways to choose  2  different colors from  4  different colors, when order is important.

Dividing by  2  to nullify the order, so after dividing order is no longer important.

2 =  Two companies for the first pen.

2=  Two companies for the second pen.

trepsixore · Answer

I have no idea if my reasoning is correct, but I got the right answer. First choosing 4C1 colors available, which is 4. Then choosing 1 out of 3 available colors left, since the colors should be different 3C1, multiplying these will give us 12. Since there are 2 brands available, multiply by 2, yield 24.

Aakanksha777 · Answer

Three scenarios

= Both pens from Company A + Both pens from Company B + One pen from A & one from B minus cases where both are same color
= 4C2 + 4C2 +  
= 6+6+ (4*4-4)
= 24

ajay0520 · Answer

what you did here worked out of luck but, for other problem it wont work. consider everything was same but number of companies were 3 instead of 2( each of 3 companies had 4 different color pen). in that case your way of approach will be 4*3*3=36. but the correct answer will be 12*9/2 =54. if you want to learn this concept then do let me know. its easy once you get the concept clear.

A shopkeeper keeps pens of two different companies Company A and Com

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A shopkeeper keeps pens of two different companies Company A and Com

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