Quant Boosters  Rajesh Balasubramanian  Set 2

Number of Questions  30
Topic  Permutation & Combination
Solved ?  Yes
Source  2IIM

Q1) John extracts three letters from the word ‘ACCEDE’ and makes words out of them, how many different words can he generate?

John can extract three distinct letters or, 2 of one kind and one different.
Scenario I: Three distinct letters
Step I: Some 3 of the 4 letters A, C, D, E can be selected. 4C3
Step II: This can be rearranged in 3! ways.
Total number of ways = 4C3 × 3! = 4 × 6 = 24.
Scenario II:
Step I
(i) 2 C’s and one of A, D or E or
(ii) 2 E’s and one of A, C or D. 6 possibilities totally
Step II: This can be rearranged in 3!/2! ways
Total number of ways = 6 × 3 = 18
24 + 18 = 42

Q2) Joseph extracts three letters from the word ‘RENEGED’ and makes words out of these 3 letters. How many such words can he generate?

Joseph can extract three distinct letters or, 2 of one kind and one different, or all 3 being the same letter.
Scenario I: Three distinct letters
Step I: Any 3 of the 5 letters R, E, N, G, D. 5C3
Step II: This can be rearranged in 3! = 6 ways.
Total number of ways = 5C3 × 3! = 10 × 6 = 60
Scenario II:
Step I: 2 E’s and one of R, G, D or N. 4 ways of selecting one of the other 4 letters.
Step II: This can be rearranged in 3!/2! = 3 ways
Total number of ways = 4 × 3 = 12
Scenario III: All three being the same letter. All three can be ‘E’. There is only one word.
60 + 12 + 1 = 73

Q3) If we listed all numbers from 100 to 10,000, how many times would the digit 3 be printed?

We need to consider all three digit and all 4digit numbers.
Three–digit numbers: A B C. 3 can be printed in the 100’s place or 10’s place or unit’s place.
100’s place: 3 B C. B can take values 0 to 9, C can take values 0 to 9. So, 3 gets printed in the 100’s place 100 times.
10’s place: A 3 C. A can take values 1 to 9, C can take values 0 to 9. So, 3 gets printed in the 10’s place 90 times.
Unit’s place: A B 3. A can take values 1 to 9, B can take values 0 to 9. So, 3 gets printed in the unit’s place 90 times.
So, 3 gets printed 280 times in 3–digit numbers.
Four–digit numbers: A B C D. 3 can be printed in the 1000’s place, 100’s place or 10’s place or unit’s place.
1000’s place: 3 B C D. B can take values 0 to 9, C can take values 0 to 9, D can take values 0 to 9. So, 3 gets printed in the 100’s place 1000 times.
100’s place: A 3 C D. A can take values 1 to 9, C & D can take values 0 to 9. So, 3 gets printed in the 100’s place 900 times.
10’s place: A B 3 D. A can take values 1 to 9, B & D can take values 0 to 9. So, 3 gets printed in the 10’s place 900 times.
Unit’s place: A B C 3. A can take values 1 to 9, B & C can take values 0 to 9. So, 3 gets printed in the unit’s place 900 times.
3 gets printed 3700 times in 4–digit numbers.
So, there are totally 3700 + 280 = 3980 numbers.
The alternative, much simpler way, would be to count the number of ways 3 would be printed while printing numbers from 1 to 10000, and then subtract the number of ways 3 would get printed from 1 to 100 from this number.
Number of ways 3 would be printed while printing numbers from 1 to 10000: 4000. 3 would get printed 1000 times in the units place, 1000 times in the tens place, 1000 times in the hundreds place and 1000 times in the thousands place.
Number of ways 3 would be printed while printing numbers from 1 to 100: 20. 10 times each in the units and tens place.
Answer = 4000  20 = 3980.
As we have mentioned before, sometimes solving by a circuitous route could be instructive. So, we will continue to take detours like these.

Q4) How many odd numbers with distinct digits can be created using the digits 1, 2, 3, 4, 5 and 6?

Single digit numbers: 1, 3 or 5: Three numbers
Two–digit numbers: Units digit = 1, 3 or 5. For the tens’ digit, there are 5 choices {anything apart from what went into the units digit}. So, there will be 3 × 5 = 15 such numbers.
Three–digit numbers: Units digit = 1, 3 or 5. For the tens’ digit, there are 5 choices {anything apart from what went into the units digit}. For the 100s’ digit, there are 4 choices {anything apart from what went into the units digit or tens digit}. So, there will be 3 × 5 × 4 = 60 such numbers.
4–digit numbers: There will be 3 × 5 × 4 × 3 = 180 numbers.
5–digit numbers: There will be 3 × 5 × 4 × 3 × 2 = 360 numbers.
6–digit numbers: There will be 3 × 5 × 4 × 3 × 2 × 1 = 360 numbers.
Total number of numbers = 360 +360 + 180 + 60 + 15 + 3 = 978.

Q5) How many 5–digit numbers with distinct digits can be created with digits 1, 2, 3, 4, 5, 6 such that the number is multiple of 12?

For a number to be a multiple of 12, it has to be a multiple of 3 and 4.
For the number to be a multiple of 3, the sum of the digits should be a multiple of 3. The sum of all 6 digits = 21. This is a multiple of 3. So, if the sum of 5 digits has to be a multiple of 3, we need to drop one multiple of 3 from this.
So, the 5 distinct digits that can go into forming the number can be 1, 2, 3, 4, 5 or 1, 2, 4, 5, 6.
Numbers with digits 1, 2, 3, 4 and 5: For the number to be a multiple of 4, the last two digits should be a multiple of 4. The last two digits can be 12, 32, 52 or 24.
When the last two digits are 12: The number is __ __ __ 12. The first 3 digits have to be 3, 4, 5 in some order. There are 3! such numbers possible. Or, there are 6 numbers in this list.
So, total number of numbers possible with the digits 1, 2, 3, 4 and 5 = 6 × 4 = 24
Numbers with digits 1, 2, 4, 5 and 6: For the number to be a multiple of 4, the last two digits should be multiple’s of 4. The last two digits can be 12, 32, 24, 64, 16 or 56.
With each of these as the last two digits, we can have 3! or 6 numbers.
So, the total number of numbers possible with the digits 1, 2, 4, 5 and 6 = 6 × 6 = 36.
The Total number of numbers = 24 + 36 = 60.

Q6) How many 6 letter words with distinct letters exist that have more vowels than consonants?

vowels and 2 consonants: 5C4 × 21C2 × 6!
5 vowels and 1 consonants: 5C5 × 21C1 × 6!
Total number of words = 5C4 × 21C2 × 6! + 5C5 × 21C1 × 6! = 6! (5 × 210 + 1 × 21) = 6! × 1071

Q7) How many 4 letters words can be created with more consonants than vowels?

4 consonants and 0 vowels: 21^4
3 consonants and 1 vowel: This can happen in three different ways
3 distinct consonants and 1 vowel: 21C3 × 5C1 × 4!
2 consonants of which 1 occurs twice and 1 vowel: 21C2 × 2C1 × 5C1 × 4!/2!
1 consonant that appears thrice and 1 vowel: 21C1 × 5C1 × 4!/3!
Total number of words = 214 + 21C3 × 5C1 × 4! + 21C2 × 2C1 × 5C1 × 4!/2! + 21C1 × 5C1 × 4!/3!
Alternatively, the above answer can be given as 214 + 213 * 20.
Try to figure out the logic behind that answer.

Q8) All the rearrangements of the word “DEMAND” are written without including any word that has 2 D’s appearing together. If these are arranged alphabetically, what would be the rank of “DEMAND”?

Number of rearrangements of word DEMAND = 6!/2! = 360
Number of rearrangements of word DEMAND where 2 D’s appear together = 5! = 120
Number of rearrangements of word DEMAND where 2D’s do not appear together = 360–120 = 240 Words starting with ‘A’; without two D’s adjacent to each other
Words starting with A: 5!/2!= 60
Words starting with A where 2 D’s are together = 4! = 24
Words starting with ‘A’, without two D’s adjacent to each other = 36  Next we have words starting with D.
Within this, we have words starting with DA: 4! words = 24 words  Then words starting with DE
Within this, words starting with DEA – 3! = 6 words  Then starting with DED – 3! = 6 words
 Then starting with DEM
First word is DEMADN
Second is DEMAND
Rank of DEMAND = 36 + 24 + 6 + 6 + 2 = 74
 Words starting with ‘A’; without two D’s adjacent to each other

Q9) If letters of the word SLEEPLESS were arranged alphabetically, what rank would ‘SLEEPLESS’ hold?

Total number of words = 9!/3!3!2! = 5040
Words starting with E = 8!/3!2!2! = 1680
Words starting with L = 8!/3!3! = 1120
Words starting with P = 8!/3!3!2! = 560
Words starting with S = 8!/3!2!2! = 1680
Words gone by thus far
Starting with E  1680
Starting with L  1120
Stating with P  560
Now, within the words starting with S
Words starting with SE = 7!/2!2!2! = 630
Words starting with SL = 7!/3!2! = 630, SLEEPLESS is within this list, so we need to drill further.
Words starting with SLEEE = 4!/2!= 12 words
Words starting with SLEEL = 4!/2! = 12 words
Words starting with SLEEPE = 3!/2! = 3 words
Words gone thus far
Starting with E  1680
Starting with L  1120
Starting with P  560
Starting with SE  630
Starting with SLEEE  12
Starting with SLEEL  12
Starting with SLEEPE  3
Then comes SLEEPLESS (at long last): Rank = 4018

Q10) If all words with 2 distinct vowels and 3 distinct consonants were listed alphabetically, what would be the rank of “ACDEF’?