Quant Boosters  Rajesh Balasubramanian  Set 1

Rearrangements of MALAYALAM = 9!/4!2!2!
Rearrangements where the A’s appear together:
Put 4 A’s together into X and rearrange MLYLMX. This can be rearranged in 6!/2!2!
Rearrangements where the consonants appear together: Put consonants together into Z. The word will be AAAAZ. This can be rearranged in 5!/4! ways.
Now, Z is MMLLY. Z can take 5!/2!2! ways.
So, total number of arrangements =5!/4! × 5!/2!2!

Q2) In Octaworld, everything is written in base 8 form. Ram’s 1050 page tome written in the real world is translated to Octa–speak and reprinted in Octa–world. In the reprint, each page number is written in base 8 form. How many times will the digit 4 be printed on the page numbers?

$(1050)_{10}$ =$(2032)_8$ .
So, we need to see how many times the digit 4 gets printed from 1 to 2032 in base 8.
Let us consider a number of the form$(abc)_8$ where a, b c take all digits from 0 to 7. Essentially, in decimal equivalent, we are considering all numbers from 1 to 511.
When c = 4, a can take all values from 0 to 7 and b can take all values from 0 to 7. So, 4 gets printed at c 64 times.
When b = 4, a can take all values from 0 to 7 and c can take all values from 0 to 7. So, 4 gets printed at b 64 times.
When a = 4, b can take all values from 0 to 7 and c can take all values from 0 to 7. So, 4 gets printed at a 64 times.
So, totally 4 gets printed 64 × 3 = 192 times from$(1)_8$ to$(777)_8$
From$(1000)_8$ to$(1777)_8$ , 4 would get printed 192 times.
So, up to$(2000)_8$ , the digit 4 gets printed 192 × 2 = 384 times.
We need to account for all base 8 numbers from$(2001)_8$ to$(2032)_8$ .
In these 26 numbers, the digit 4 gets printed thrice$(2004)_8$ ,$(2014)_8$ ,$(2024)_8$ .
So, the digit 4 gets printed 384 + 3 = 387 times.

Q3) When a die is thrown twice, in how many outcomes will the product of the two throws be 12?

12 can be formed from 3 × 4 or 2 × 6 {1 × 12 is not possible with die}.
This can happen in 2 + 2 = 4 ways.

Q4) How many words exist that have exactly 5 distinct consonants and 2 vowels?

Scenario I: 5 distinct consonants and 2 distinct vowels – Number of words = 21C5 × 5C2 × 7!
Scenario II: 5 distinct consonants and 1 vowel appearing twice – Number of words = 21C5 × 5C1 × 7!/2!

Q5) When a coin is tossed 6 times, in how many outcomes will there be more heads than tails?

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 =$2^3$ .
Of these there is one option where no CD is picked. We need to exclude that option.So, number of possibilities =
$2^3$ – 1
Total number of outcomes = 4 × ($2^3$ – 1)
= 4 × 7
= 28

Q8) What is sum of all 4digit 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 =
$2^{10}$
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$2^{10}$ subsets. We should bear in mind that this$2^{10}$ includes the 2 improper subsets as well. The whole set P and the null set are included in this$2^{10}$ .
Subsets whose product is not a multiple of 3 = Subsets of the set {1, 2, 4, 5, 7, 8, 10} =$2^{7}$ . This includes the empty subset also. So, the correct answer should be$2^{7}$ –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?