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Bunuel
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Tip: You use combination when the question asks you to find unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as: nCk (read as From n objects choose k objects)

Tip 2: Simultaneous events are always multiplied


\(4C2*5C4*3C2=6*5*3=90\)


Isn't 4C2 = 12, not 6?

Can you please explain?

nCk = n!/[(k!)*(n-k)!]

So in this case, 4C2 = 4!/[(2!)*(2!)] = 3! (because 2!*2! = 4, so 4! becomes 3!) = 6.

I hope that helps.
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Tip: You use combination when the question asks you to find unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as: nCk (read as From n objects choose k objects)

Tip 2: Simultaneous events are always multiplied


\(4C2*5C4*3C2=6*5*3=90\)


Isn't 4C2 = 12, not 6?

Can you please explain?


The Formula for combination is \(nCr = \frac{n!}{(n-r)*r!]}\)
\(4C2 = \frac{4!}{[(4-2!)*(2!)]} = \frac{4*3*2*1}{2*1*2*1}\)


Ive been using this link this to learn...it did help me... you might want to take a look.

https://gmatclub.com/forum/combination- ... 56486.html
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We can use combinations to solve this

4C2*5C4*3C2

=6*5*3
=90

Option D
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Bunuel
A stock broker recommends a portfolio of 2 Information Technology stocks, 4 Retail stocks, and 2 e-commerce stocks. If the broker can choose from 4 Information Technology stocks, 5 Retail stocks, and 3 e-commerce stocks, how many different portfolios of 8 stocks are possible?

(A) 9
(B) 24
(C) 60
(D) 90
(E) 120

The number of ways to select the information technology stocks is 4C2:

4C2 = (4 x 3)/2! = 6

The number of ways to select the retail stocks is 5C4:

5C4 = 5

The number of ways to select the e-commerce stocks is 3C2:

3C2 = 3

So the total number of ways of selecting the portfolio of 8 stocks is 6 x 5 x 3 = 90.

Answer: D
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