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Out of seven different pens that I have, each day I take two pens to t
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16 Nov 2021, 05:27
16 Nov 2021, 05:27
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Out of seven different pens that I have, each day I take two pens to the office. If the combinations of the two pens on any given day is not the same as that on any other day, find the numbers of days that are required to exhaust all such combinations.
(A) 7
(B) 14
(C) 20
(D) 21
(E) 22
(A) 7
(B) 14
(C) 20
(D) 21
(E) 22
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Re: Out of seven different pens that I have, each day I take two pens to t
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16 Nov 2021, 06:01
16 Nov 2021, 06:01
1
Carcass wrote:
Out of seven different pens that I have, each day I take two pens to the office. If the combinations of the two pens on any given day is not the same as that on any other day, find the numbers of days that are required to exhaust all such combinations.
(A) 7
(B) 14
(C) 20
(D) 21
(E) 22
(A) 7
(B) 14
(C) 20
(D) 21
(E) 22
In other words, we want to determine the total number of ways to select 2 pens out of 7 pens.
Since the order in which we select the pens does not matter, we can use combinations.
We can select 2 pens out of 7 pens in 7C2 ways (= 21 ways)
So, we can bring a different pair of pens each day for 21 days.
Answer: D
ASIDE: Here's a video lesson on quickly calculating combinations (e.g., 7C2) in your head:
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Re: Out of seven different pens that I have, each day I take two pens to t
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16 Nov 2021, 08:08
16 Nov 2021, 08:08
1
From my perspective, there's no point in bringing out the tired n!/(r!(n-r)!) equations, so let's leave those to the side. Remember, equations often cloud true understanding so use with care (or don't, in this case 
).
You can answer any P&C question (this is a C question, for the record) using three questions:
1) How Many Spaces?
2) How Many Choices?
3) Does Order Matter?
First, the number of spaces we have is two because we're taking two pens.
_ _
Second, the number of choices would be the pool from which we choose, or seven.
7*6
Third, does order matter? No, because "Pen A then Pen B" would be the same two pens as "Pen B then Pen A." If the order does not matter*, divide by factorial of the group size (number of spaces). In this case, that means 2!.
7*6 / 2!
That gives us 42/2! = 42/2 = 21. The answer is D.
*In cases where the order does matter, as in "Pen A then Pen B" would be treated as different from "Pen B then Pen A," simply omit division by the factorial here. That's a P rather than a C.
).You can answer any P&C question (this is a C question, for the record) using three questions:
1) How Many Spaces?
2) How Many Choices?
3) Does Order Matter?
First, the number of spaces we have is two because we're taking two pens.
_ _
Second, the number of choices would be the pool from which we choose, or seven.
7*6
Third, does order matter? No, because "Pen A then Pen B" would be the same two pens as "Pen B then Pen A." If the order does not matter*, divide by factorial of the group size (number of spaces). In this case, that means 2!.
7*6 / 2!
That gives us 42/2! = 42/2 = 21. The answer is D.
*In cases where the order does matter, as in "Pen A then Pen B" would be treated as different from "Pen B then Pen A," simply omit division by the factorial here. That's a P rather than a C.
