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Intern  Joined: 01 Dec 2012
Posts: 36
Location: India
WE: Project Management (Retail)
A security agency assigns a two-digit code with distinct dig  [#permalink]

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Pemutations & Combinations Questions

Q1. A security agency assigns a two-digit code with distinct digits to each of its members using the digits 0,1,2,3....,9 such that the first digit of the code is not zero. However the code printed on a badge can potentially create confusion when read upside down. For example, the code 18 may appear as 81. How many codes are there for which there for which no such confusion can arise?
a. 69 b. 71 c. 81 d. 65 e. 68

Q2. Find the number of ways of dividing 16 different books equally among
(i) 4 boys?
a. $$\frac{16!}{(4!)^4}$$ b. $$\frac{16!}{(4!)^3}$$ c. $$\frac{16!}{(4!)^5}$$ d. $$\frac{16!}{4!}$$e. $$(4!)^4$$

(ii) into 4 parcels?
a. $$\frac{16!}{(4!)^4}$$ b. $$\frac{16!}{(4!)^3}$$ c.$$\frac{16!}{(4!)^5}$$ d. $$\frac{16!}{4!}$$ e. $$(4!)^4$$

Q3. Of twelve points on a plane, 5 points lie on a straight line and no other combination of 3 points lie on a straight line. Then, by joining these points how many quadrilaterals can be formed?
a. 495 b. 460 c. 490 d. 480 e. 420

Q4. Let S be a set of all 5-digit numbers with distinct digits that can be formed using the digits 1,2,4,5 and 8 such that exactly two odd positions are occupied by the even digits. Find the sum of the digits in the rightmost position of all the numbers in S.
a. 296 b. 256 c. 316 d. 269 e. 329

Q5. Ten points are plotted on a plane such that no three of them lie on a straight line. Four of these points are joined to each of the remaining six points and each of the remaining six points is joined to exactly five points. How many line segments are formed?
a. 28 b. 25 c. 29 d. 27 e. 24

Q6. Using the first ten letters of the english alphabet how many strings of four letters can be formed such that each H is followed by B?
a. 6760 b. 6516 c. 6561 d. 240 e. 6805

Q7. Mac has forgotten his friends 8-digit telephone number but remembers the following:
i) the first 3 digit are either 270 or 279.
ii) the digit 0 occurs exactly three times and the digit 9 occurs exactly one time.
iii) the number was an even number.
If he were to use a trial & error method to reach his friend, what is the minimum number of trials he has to make to be sure to succeed?
a. 1664 b. 1544 c. 832 d. 1280 e. 2000

Q8. Two red, three balck & two white pencils are to be arrange in a row such that:
i) no two adjacent pencils are of the same color &
ii) the pencils at the two ends of the row are of same color.
In how many ways can the pencils be arranged?
a. 12 b. 8 c. 10 d. 9 e. 11

Any shortcut for the above questions...??? pls help!!

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Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9647
Location: Pune, India
Re: A security agency assigns a two-digit code with distinct dig  [#permalink]

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pariearth wrote:
Pemutations & Combinations Questions

Q1. A security agency assigns a two-digit code with distinct digits to each of its members using the digits 0,1,2,3....,9 such that the first digit of the code is not zero. However the code printed on a badge can potentially create confusion when read upside down. For example, the code 18 may appear as 81. How many codes are there for which there for which no such confusion can arise?
a. 69 b. 71 c. 81 d. 65 e. 68

First think of how the numbers change when the card is held upside down. The digit on the right comes to the left and the digit on the left goes to the right.

18 -> 81

Also the numbers turn upside down. Since 1 and 8 are symmetrical, turning them upside down doesn't change anything.
Now, 0, 1 and 8 are symmetrical. When you turn 6 upside down, it becomes 9 so it is a source of confusion too.
So how many numbers can create confusion? Since 0 cannot be on the left side in the number, a number that has 0 cannot create confusion (even if the 0 is on the right side, when you turn the card, it will come to the left which is not acceptable)
So the tens digit can take one of 4 values (1, 6, 8, 9) and the unit's digit can take one of the leftover 3 values in 4*3 = 12 ways
This 12 includes 69 and 96. Note that when you turn 69 upside down, it stays 69 (9 comes to the left and 6 goes to the right but 9 becomes 6 and 6 becomes 9)
Hence the numbers that can cause confusion = 12 - 2 = 10
Total numbers that can be formed = 9*9 = 81 (select tens digit in 9 ways and units in 9 ways)
No of numbers in which no confusion arises = 81 - 10 = 71
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Intern  Joined: 06 Dec 2012
Posts: 23
Concentration: Finance, International Business
GMAT 1: 510 Q46 V21 GPA: 3.5
Re: A security agency assigns a two-digit code with distinct dig  [#permalink]

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pariearth wrote:
Pemutations & Combinations Questions

Q1. A security agency assigns a two-digit code with distinct digits to each of its members using the digits 0,1,2,3....,9 such that the first digit of the code is not zero. However the code printed on a badge can potentially create confusion when read upside down. For example, the code 18 may appear as 81. How many codes are there for which there for which no such confusion can arise?
a. 69 b. 71 c. 81 d. 65 e. 68

Q2. Find the number of ways of dividing 16 different books equally among
(i) 4 boys?
a. $$\frac{16!}{(4!)^4}$$ b. $$\frac{16!}{(4!)^3}$$ c. $$\frac{16!}{(4!)^5}$$ d. $$\frac{16!}{4!}$$e. $$(4!)^4$$

(ii) into 4 parcels?
a. $$\frac{16!}{(4!)^4}$$ b. $$\frac{16!}{(4!)^3}$$ c.$$\frac{16!}{(4!)^5}$$ d. $$\frac{16!}{4!}$$ e. $$(4!)^4$$

Q3. Of twelve points on a plane, 5 points lie on a straight line and no other combination of 3 points lie on a straight line. Then, by joining these points how many quadrilaterals can be formed?
a. 495 b. 460 c. 490 d. 480 e. 420

Q4. Let S be a set of all 5-digit numbers with distinct digits that can be formed using the digits 1,2,4,5 and 8 such that exactly two odd positions are occupied by the even digits. Find the sum of the digits in the rightmost position of all the numbers in S.
a. 296 b. 256 c. 316 d. 269 e. 329

Q5. Ten points are plotted on a plane such that no three of them lie on a straight line. Four of these points are joined to each of the remaining six points and each of the remaining six points is joined to exactly five points. How many line segments are formed?
a. 28 b. 25 c. 29 d. 27 e. 24

Q6. Using the first ten letters of the english alphabet how many strings of four letters can be formed such that each H is followed by B?
a. 6760 b. 6516 c. 6561 d. 240 e. 6805

Q7. Mac has forgotten his friends 8-digit telephone number but remembers the following:
i) the first 3 digit are either 270 or 279.
ii) the digit 0 occurs exactly three times and the digit 9 occurs exactly one time.
iii) the number was an even number.
If he were to use a trial & error method to reach his friend, what is the minimum number of trials he has to make to be sure to succeed?
a. 1664 b. 1544 c. 832 d. 1280 e. 2000

Q8. Two red, three balck & two white pencils are to be arrange in a row such that:
i) no two adjacent pencils are of the same color &
ii) the pencils at the two ends of the row are of same color.
In how many ways can the pencils be arranged?
a. 12 b. 8 c. 10 d. 9 e. 11

Any shortcut for the above questions...??? pls help!!

can anyone help solving 7th question ?
Intern  Joined: 06 Dec 2012
Posts: 23
Concentration: Finance, International Business
GMAT 1: 510 Q46 V21 GPA: 3.5
Re: A security agency assigns a two-digit code with distinct dig  [#permalink]

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VeritasPrepKarishma wrote:
pariearth wrote:
Pemutations & Combinations Questions

Q1. A security agency assigns a two-digit code with distinct digits to each of its members using the digits 0,1,2,3....,9 such that the first digit of the code is not zero. However the code printed on a badge can potentially create confusion when read upside down. For example, the code 18 may appear as 81. How many codes are there for which there for which no such confusion can arise?
a. 69 b. 71 c. 81 d. 65 e. 68

First think of how the numbers change when the card is held upside down. The digit on the right comes to the left and the digit on the left goes to the right.

18 -> 81

Also the numbers turn upside down. Since 1 and 8 are symmetrical, turning them upside down doesn't change anything.
Now, 0, 1 and 8 are symmetrical. When you turn 6 upside down, it becomes 9 so it is a source of confusion too.
So how many numbers can create confusion? Since 0 cannot be on the left side in the number, a number that has 0 cannot create confusion (even if the 0 is on the right side, when you turn the card, it will come to the left which is not acceptable)
So the tens digit can take one of 4 values (1, 6, 8, 9) and the unit's digit can take one of the leftover 3 values in 4*3 = 12 ways
This 12 includes 69 and 96. Note that when you turn 69 upside down, it stays 69 (9 comes to the left and 6 goes to the right but 9 becomes 6 and 6 becomes 9)
Hence the numbers that can cause confusion = 12 - 2 = 10
Total numbers that can be formed = 9*9 = 81 (select tens digit in 9 ways and units in 9 ways)
No of numbers in which no confusion arises = 81 - 10 = 71

Need help on solving 7th question...
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1835
Re: A security agency assigns a two-digit code with distinct dig  [#permalink]

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pariearth wrote:

Q7. Mac has forgotten his friends 8-digit telephone number but remembers the following:
i) the first 3 digit are either 270 or 279.
ii) the digit 0 occurs exactly three times and the digit 9 occurs exactly one time.
iii) the number was an even number.
If he were to use a trial & error method to reach his friend, what is the minimum number of trials he has to make to be sure to succeed?
a. 1664 b. 1544 c. 832 d. 1280 e. 2000

These questions are not at all similar to real GMAT questions - they're way too complicated, and there aren't any real shortcuts. In this phone number question, we have four separate cases:

Case 1: phone number is 270-XXXX-0

- one of the four X's is a nine, so we have 4 choices for where to put the nine
- then one of the three remaining X's is a zero, so we have 3 choices for where to put the zero
- then we have 8 choices for the two remaining digits (they cannot be 0 or 9)

So we have 4*3*8*8 possibilities

Case 2: phone number is 270-XXXX-E, where E is even and nonzero

- we have 4 choices for E (it is 2, 4, 6, or 8)
- one of the four X's is a nine, so we have 4 choices
- two of the remaining three digits are zero, so we have 3 choices (3C2)
- for the remaining digit we have 8 choices (cannot be 0 or 9)

So we have 4*4*3*8 possibilities

Case 3: phone number is 279-XXXX-0

- we have 4C2 = 6 choices for where to put the remaining zeros
- we have 8 choices for each of the two other digits

So we have 6*8*8 possibilities

Case 4: phone number is 279-XXXX-E where E is even and nonzero

- we have 4 choices for E
- we have 4C3 = 4 choices for where to put the three zeros
- we have 8 choices for the other digit

So we have 4*4*8 possibilities.

Adding the possibilities from each case, we get the answer:

4*3*8*8 + 4*4*3*8 + 6 * 8 * 8 + 4 * 4 * 8 = 1664

I guarantee you will never see a problem anywhere close to as awkward as this on the GMAT - when you have cases in a GMAT counting problem and no shortcut that allows you to bypass the case analysis, you will almost always have only two cases to consider.

--== Message from the GMAT Club Team ==--

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This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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