Quant Boosters - Rajesh Balasubramanian - Set 1



  • We should have more heads than tails => There should be 4 heads or 5 heads or all 6 heads.
    Number of ways = 6C4 + 6C5 + 6C6
    = 15 + 6 + 1
    = 22



  • Q6) In how many ways can we pick 4 cards from a card pack such that there are no Aces selected and there are more face cards than numbered cards?



  • Scenario I: 3 face cards and 1 numbered card: 12C3 × 36C1
    Scenario II: 4 face cards and 0 numbered cards: 12C4
    Therefore, total number of ways is 12C3 × 36C1 + 12C4



  • Q7) On a table, there are 4 identical copies of a book and 3 CDs. In how many ways can we pick at least one book and at least one CD from the table?



  • There are 4 identical copies of a book. One can pick either 0, 1, 2, 3, or all 4 of these – 5 different options. We need to pick at least one book. So, we have only 4 options – 1, 2, 3, or 4 books being picked
    There are 3 different CDs. Each CD can be either picked or not picked. So, total number of options = 232^3.
    Of these there is one option where no CD is picked. We need to exclude that option.

    So, number of possibilities = 232^3 – 1
    Total number of outcomes = 4 × ( 232^3 – 1)
    = 4 × 7
    = 28



  • Q8) What is sum of all 4-digit numbers formed by rearranging the digits of the number 2235?



  • Number of rearrangements of 2235 = 4!/2! = 12.
    So, we need to add these 12 numbers.
    Let us consider the units’ digit of these 12 numbers.
    The units digit will be the one of the digits 2, 3, or 5. If the last digit were 3, the first 3 digits should be some rearrangement of 2, 2, and 5. So, there are 3!/2! Such numbers. Or, 3 such numbers.
    Similarly there are three numbers with 5 as the units digit.
    If the last digit were 2, the first 3 digits should be some rearrangement of 2, 3, and 5. So, there are 3! such numbers, or, 6 such numbers.
    So, the units digit will be 2 for six numbers, 3 for three numbers, and 5 for three numbers. Sum of all these unit digits will be 2 × 6 + 3 × 3 + 5 × 3 = 12 + 9 + 15 = 36.
    Sum of all the tens digits will be 36. Sum of all the digits in the hundreds’ place will be 36.
    Sum of all the digits in the thousands’ place is 36.
    So, sum of all the 4–digit numbers will be 36 × 1111 = 39996.



  • Q9) When a die is thrown twice, in how many ways can we have the sum of numbers to be less than 8?



  • Sum of the numbers seen in the two throws can be 2, 3, 4, 5, 6 or 7.
    Sum of the digits = 2: This can only be 1 + 1. One way
    Sum of the digits = 3: This can be 1 + 2 or 2 + 1. 2 ways
    Sum of the digits = 4: 1 + 3, 3 + 1, 2 + 2. 3 ways
    Sum of the digits = 5: 1 + 4, 4 + 1, 3 + 2, 2 + 3. 4 ways
    Sum of the digits = 6: 1 + 5, 5 + 1, 2 + 4, 4 + 2, 3 + 3. 5 ways
    Sum of the digits = 7: 1 + 6, 6 + 1, 2 + 5, 5 + 2, 3 + 4, 4 + 3. 6 ways
    Sum of the numbers in the two throws can be less than 8 in 1 + 2 + 3 + 4 + 5 + 6 = 21 ways.
    We notice a very simple pattern here. Try the sum of the numbers all the way to 12 and see the rest of the pattern also.



  • Q10) Set P has elements {1, 2, 3..…10}. How many non–empty subsets of P have the product of their elements as not a multiple of 3?



  • Total number of subsets = 2102^{10}
    For choosing any subset, each element can either be part of the subset or not part of the subset. So, for each element, there are two options. So, with 10 elements in the set, we can create 2102^{10} subsets. We should bear in mind that this 2102^{10} includes the 2 improper subsets as well. The whole set P and the null set are included in this 2102^{10}.
    Subsets whose product is not a multiple of 3 = Subsets of the set {1, 2, 4, 5, 7, 8, 10} = 272^{7}. This includes the empty subset also. So, the correct answer should be 272^{7} –1



  • Q11) A, B, C, D, E are doctors, P, Q, R, S, are engineers. In how many ways can we select a committee of 5 that has more engineers than doctors?



  • Two scenarios are possible.
    3 engineers and 2 doctors: 4C3 × 5C2 = 4 × 10 = 40
    4 engineers and 1 doctor : 4C4 × 5C1 = 1 × 5 = 5
    Total number of possibilities = 40 + 5 = 45



  • Q12) From a card pack of 52, in how many ways can we pick a sequence of 4 cards such that they are in order and from different suits? Consider Ace to be the card following King in each suit. So, Ace can be taken to precede ‘2’ and succeed ‘King’. So, JQKA would be a sequence, so would be A234. However, QKA2 is not a sequence.



  • 4 cards in order can be A234, 2345, ….JQKA. 11 different possibilities
    For a given set of four cards, say 2345, they can be from 4 different suits in 4! ways.
    So, total number of possibilities = 11 × 4! = 264.



  • Q13) In how many ways can letters of the word ATTITUDE be rearranged such that no two T’s are adjacent to each other?



  • ATTITUDE has 8 letters, of which 3 are T’s
    Now, Let us place the letters that are not Ts on a straight line. We have AIUDE. These can be arranged in 5! Ways. Now let us create slots between these letters to place the Ts in. In order to ensure that no two T’s are adjacent to each other, let us create exactly one slot between any two letters.
    A __ I __ U __ D __ E.
    Additionally, let us add one slot at the beginning and end as well as the T’s can go there also
    A __ I __ U __ D __ E
    Now, out of these 6 slots, some 3 can be T. That can be selected in 6C3 ways.
    So, total number of words = 5! × 6C3 = 2400



  • Q14) In how many rearrangements of the word SLEEPLESS will no two S’s appear together?



  • Let us aggregate the other letters LEEPLE and arrange these.
    These can be arranged in 6!/2!3! ways. Now, let us create slots in between and before/after these letters, where S can potentially appear.
    __ L __ E __ E __ P __ L __ E__.
    Out of these 7 slots, 3 should be taken up by S’s. This can be done in 7C3 ways.
    So, total number of rearrangements = 6!/2!3! × 7C3 = 60 × 35 = 2100



  • Q15) How many numbers of up to 5 digits can be created using the digits 1, 2 and 3 that are multiples of 12?


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