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Bunuel
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Simon has P different toys in a bag. If he discards Q toys, then in 09 Sep 2024, 01:30
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85% (hard)

Question Stats:

44% (01:46) correct 56% (02:01) wrong based on 213 sessions

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­Simon has P different toys in a bag. If he discards Q toys, then in how many ways he can select 8 toys from the remaining toys in the bag?

(1) If Simon had discarded 3 more toys from the bag, he could have made 495 selections.
(2) P = Q + 15


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Simon has P different toys in a bag. If he discards Q toys, then in 09 Sep 2024, 02:43
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Total toys= P
Discarded toys=Q
Remaining toys= P-Q
To find (P-Q)C8 = ?

1) (P-Q-3)C8= 495 [ 12C8= 495]
Hence, P-Q-3= 12
P-Q= 15

Sufficient

2) P=15+Q
P-Q=15

Sufficient


Either A or B is sufficient to answer the question
Hence , correct answer is D.

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Simon has P different toys in a bag. If he discards Q toys, then in 09 Sep 2024, 08:37
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Ayeka
Total toys= P
Discarded toys=Q
Remaining toys= P-Q
To find (P-Q)C8 = ?

1) (P-Q-3)C8= 495 [ 12C8= 495]
Hence, P-Q-3= 12
P-Q= 15

Sufficient

2) P=15+Q
P-Q=15

Sufficient


Either A or B is sufficient to answer the question
Hence , correct answer is D.

Posted from my mobile device
what's the way to come up with ­this (495 = 12C8) though? Any trick or efficient method?
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Simon has P different toys in a bag. If he discards Q toys, then in 09 Sep 2024, 12:10
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Ayeka
Total toys= P
Discarded toys=Q
Remaining toys= P-Q
To find (P-Q)C8 = ?

1) (P-Q-3)C8= 495 [ 12C8= 495]
Hence, P-Q-3= 12
P-Q= 15

Sufficient

2) P=15+Q
P-Q=15

Sufficient


Either A or B is sufficient to answer the question
Hence , correct answer is D.

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­Same as above how did you come to this as 12C8
its fascinating
coz I felt no not solvable and moved on
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Simon has P different toys in a bag. If he discards Q toys, then in 12 Sep 2024, 03:08
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EshaFatim
Ayeka
Total toys= P
Discarded toys=Q
Remaining toys= P-Q
To find (P-Q)C8 = ?

1) (P-Q-3)C8= 495 [ 12C8= 495]
Hence, P-Q-3= 12
P-Q= 15

Sufficient

2) P=15+Q
P-Q=15

Sufficient


Either A or B is sufficient to answer the question
Hence , correct answer is D.

Posted from my mobile device
what's the way to come up with ­this (495 = 12C8) though? Any trick or efficient method?


Checking by approximation where xC8=495 might fall. 495 isn’t too big so it has to fall 3-5 points away from 8. Going by 9C8=9, 10C8=45, 11C8=265, 12C8=495

Also, in most DS questions the hint to solve xC8 is already given either in the question or other option. It’s only 700+ questions where you might not be given any hint to solve it, and has to use approximation.
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Simon has P different toys in a bag. If he discards Q toys, then in 12 Sep 2024, 22:09
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Shwarma
Ayeka
Total toys= P
Discarded toys=Q
Remaining toys= P-Q
To find (P-Q)C8 = ?

1) (P-Q-3)C8= 495 [ 12C8= 495]
Hence, P-Q-3= 12
P-Q= 15

Sufficient

2) P=15+Q
P-Q=15

Sufficient


Either A or B is sufficient to answer the question
Hence , correct answer is D.

Posted from my mobile device
­Same as above how did you come to this as 12C8
its fascinating
coz I felt no not solvable and moved on


Checking by approximation where xC8=495 might fall. 495 isn’t too big so it has to fall 3-5 points away from 8. Going by 9C8=9, 10C8=45, 11C8=265, 12C8=495

Also, in most DS questions the hint to solve xC8 is already given either in the question or other option. It’s only 700+ questions where you might not be given any hint to solve it, and has to use approximation.
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Simon has P different toys in a bag. If he discards Q toys, then in 18 Sep 2024, 13:10
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Might be a stupid question, but for statement 1 ( (1) If Simon had discarded 3 more toys from the bag, he could have made 495 selections), why do we assume k = 8 in n(C)k..
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Simon has P different toys in a bag. If he discards Q toys, then in 18 Sep 2024, 21:11
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shaurya_gmat
Might be a stupid question, but for statement 1 ( (1) If Simon had discarded 3 more toys from the bag, he could have made 495 selections), why do we assume k = 8 in n(C)k..

K=8 is given in the question (we have to find the number of ways (n) in which he can select (k) 8 toys.
­Simon has P different toys in a bag. If he discards Q toys, then in how many ways he can select 8 toys from the remaining toys in the bag?

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Simon has P different toys in a bag. If he discards Q toys, then in 03 Nov 2024, 00:06
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Encountered a quick tip/concept to solve such questions.
If you're ever given nCr = X, where X and r are constants, there is bound to be a unique value for n that satisfies the condition. In the above question, we need to find (P-Q)C8. Effectively, we only need to find P-Q.
Statement 1 says that (P-Q-3)C8=X, where X is a defined constant. That means P-Q-3 will be a unique value.
We can say P-Q-3 = Y (where Y is some defined constant)
From the above, we can say P-Q = Y+3
Since we know Y, we know Y+3. Since we know Y+3, we know P-Q.
We don't need to get the exact values by guessing for this question, just knowing you will get a definitive value is more than sufficient.
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