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Permutaion and Combination- Books arrangemnt

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Combinations

by Quasar Chunawala » Fri May 02, 2014 11:59 pm
B. No. of ways to select 5 out of 12 books(such that if A is selected, C also must be selected)
(i) Suppose A is selected, then C is also selected.
C(10,3) ways
(ii) Suppose A is not selected, then C is also not selected.
C(10,5) ways
Total number of ways = C(10,3) + C(10,5) = 120 + 252 = 372

Not sure however, if my answer for (A) tallies, with what is mentioned.
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by s91arvindh » Sat May 03, 2014 2:19 am
Hi Quasar,

Thanks a lot for heads up :)
There are 12 books - A, B, C,......K, L in a shelf. In how many ways can 5 books be selected?

A) A and C cannot be selected together
B) If A is selected, then C must also be selected
For A:

Case 1: With A selected there are 11c4 ways which is 330
Case 2: With C selected there are 11c4 ways which is 330
Case 3: With A and C not selected there are 12C5 ways which is 252

Hence [spoiler]Case1+Case2+Case3 = 912[/spoiler]

For B:

Case 1: With A and C must be selected there are 10c3 ways which is 120
Case 2: With A and C not selected there are 10c5 ways which is 252

Hence [spoiler]Case1+Case2 = 372[/spoiler]

Got the answer after trying the same questions 15 minutes later :D
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by GMATGuruNY » Sat May 03, 2014 2:49 am
s91arvindh wrote:There are 12 books - A, B, C,......K, L in a shelf. In how many ways can 5 books be selected?

Q1: A and C cannot be selected together
Q2: If A is selected, then C must also be selected

Q1: A and C cannot both be selected
GOOD combinations = (ALL possible combinations - (BAD combinations with both A and C)

ALL:
Number of ways to choose 5 books from 12 options = 12C5 = (12*11*10*9*8)/(5*4*3*2*1) = 11*9*8 = 792.
BAD:
In a bad combination, 3 books are chosen to be combined with A and C.
From the 10 remaining books, the number of ways to choose 3 to be combined with A and C = 10C3 = (10*9*8)/(3*2*1) = 120.
GOOD:
792-120 = 672.

Q2: If A is selected, C is also selected
Case 1: A and C are both selected
As shown above, the number of ways to choose 3 books to be combined with A and C = 120.

Case 2: A is not selected
If A is not selected, it is still possible to select C.
From the 11 remaining books, the number of ways to choose 5 = (11*10*9*8*7)/(5*4*3*2*1) = 462.

Total ways = 120 + 462 = 582.

The OAs are incorrect.
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by GMATGuruNY » Sat May 03, 2014 2:55 am
Please see my notes in red:
s91arvindh wrote:
There are 12 books - A, B, C,......K, L in a shelf. In how many ways can 5 books be selected?

A) A and C cannot be selected together
B) If A is selected, then C must also be selected
For A:

Case 1: With A selected there are 11c4 ways which is 330
Once A is chosen, we cannot choose C.
Thus, 10 books remain that could be combined with A.
From the 10 remaining books, the number of ways to choose 4 = 10C4 = 210.

Case 2: With C selected there are 11c4 ways which is 330
Once C is chosen, we cannot choose A.
Thus, 10 books remain that could be combined with C.
From the 10 remaining books, the number of ways to choose 4 = 10C4 = 210.

Case 3: With A and C not selected there are 10C5 ways which is 252

Hence Case1+Case2+Case3 = 210+210+252 = 672.

For B:

Case 1: With A and C must be selected there are 10c3 ways which is 120
Case 2: With A and C not selected there are 10c5 ways which is 252
As I noted in my solution above, if A is not selected, we may still select C.
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by Quasar Chunawala » Sat May 03, 2014 3:34 am
GmatGuruNY,

For (A), in fact I deduced that
No. of ways of (not selecting A and C together) = Total - (no. of ways to have both A and C)
But, the answer caused doubt. You offered a very lucid explanation.

And that C may be selected, even if A isn't. Nice catch! I wish I'd have spotted that. Gee thanks!

I am a starter, I plan to take my GMAT soon. Look forward to a lot more fun...
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