Quant Boosters - Rajesh Balasubramanian - Set 2


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    We need to consider all three digit and all 4-digit 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.


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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.


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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?


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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.


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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.


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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”?


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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

    1. 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
    2. Next we have words starting with D.
      Within this, we have words starting with DA: 4! words = 24 words
    3. Then words starting with DE
      Within this, words starting with DEA – 3! = 6 words
    4. Then starting with DED – 3! = 6 words
    5. Then starting with DEM
      First word is DEMADN
      Second is DEMAND

    Rank of DEMAND = 36 + 24 + 6 + 6 + 2 = 74


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


    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


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012, 2014 and 2017.


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


 

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