Quiz 3

One of our facvourite questions. One of those that get simplified if we think about it the right way, but is extremely tough if you do not get the right line of thought

a + b + c = 10. Now, let us place ten sticks in a row

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This question now becomes the equivalent of placing two symbols somewhere between these sticks. For instance,
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This would be the equivalent of 4 + 5 + 1. or, a = 4, b = 5, c = 1.
There are 9 slots between the sticks, out of which one has to select 2 for placing the s.

The number of ways of doing this would be ⁹C₂. Bear in mind that this kind of calculation counts ordered triplets. (4, 5, 1) and (1, 4, 5) will both be counted as distinct possibilities.
We can also do a + b + c = n where a, b, c have to be whole numbers (instead of natural numbers as in this question) with a small change to the above approach. Give it some thought.

The question is "Sum of three Natural numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?"

Hence the answer is "⁹C₂ = 36."

a + b + c = 10. a, b, c are whole numbers. Now this is similar to the previous question that we solved by placing 10 sticks and simplifying.

We cannot follow an exactly similar approach, as in this case a, b and c can be zero. Let us modify the approach a little bit. Let us see if we can remove the constraint that a, b, c can be zero.

If we give a minimum of 1 to a, b, c then the original approach can be used. And then we can finally remove 1 from each of a, b, c. So, let us distribute 13 sticks across a, b and c and finally remove one from each.
a + b + c = 13. Now, let us place ten sticks in a row

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This question now becomes the equivalent of placing two \'+\' symbols somewhere between these sticks. For instance,
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This would be the equivalent of 4 + 5 + 4. or, a = 4, b = 5, c = 4.
There are 12 slots between the sticks, out of which one has to select 2 for placing the \'+\'s.
The number of ways of doing this is ¹²C₂.
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The question is "Sum of three Whole numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?"

Hence the answer is "66"

Another wonderful question. A classic question that is from a standard template once a layer is removed.

This is nothing but the number of ways of having a, b, c such that a + b + c = 11, where a, b, c are natural numbers. By having them to be natural numbers, we ensure that no box can be empty. (no zeroes)

The question is similar to the previous one. ¹⁰C₂ = 45 ways

The question is "In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is empty?"

Hence the answer is "45"

Choice C is the correct answer.

Exactly one of ab, bc and ca is odd => Two are odd and one is even.

abc is a multiple of 4 => the even number is a multiple of 4.
The arithmetic mean of a and b is an integer => a and b are odd.
and so is the arithmetic mean of a, b and c. => a + b + c is a multiple of 3.

c can be 4 or 8.
c = 4; a, b can be 3, 5 or 5, 9
c = 8; a, b can be 3, 7 or 7, 9
Four triplets are possible.

The question is "How many such triplets are possible (unordered triplets)?"

Hence the answer is "4"

Number should be a multiple of 3 and 4. So, the sum of the digits should be a multiple of 3. We can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3.
(The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.
All seven 3's - No possibility.

Three 2's and four 3's - The first 5 digits should have two 2\'s and three 3's in some order.
No of possibilities = 5!3!2!5!3!2! = 10

Six 2\'s and one 3 - The first 5 digits should all be 2\'s. So, there is only one number 2222232.
So, there are a total of 10 + 1 = 11 solutions.

The question is "A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12?"

Hence the answer is "11"

The first word would be ABCDE. With 2 distinct vowels, 3 distinct consonants, this is the first word we can come up with.

Starting with AB, we can have a number of words.

AB __ __ __. The next three slots should have 2 consonants and one vowel. This can be selected in ²⁰C₂ and ⁴C₁ ways. Then the three distinct letters can be rearranged in 3! Ways.
Or, number of words starting with AB = ²⁰C₂ * ⁴C₁ * 3! = 190 * 4 * 6 = 4560

Next, we move on to words starting with ACB
ACB __ __. The last two slots have to be filled with one vowel and one consonant. = ¹⁹C₁ * ⁴C₁. This can be rearranged in 2! Ways.
Or, number of words starting with ACB = ¹⁹C₁ * ⁴C₁ * 2 = 19 * 4 * 2 = 152

Next we move on words starting with ACDB: There are 4 different words on this list – ACDBE, ACDBI, ACDBO, ACDBU
So far number of words gone = 4560 + 152 + 4 = 4716

Starting with AB          4560
Starting with ACB         152
Starting with ACDB          4
Total words gone        4716

After this we move to words starting with ACDE, the first possible word is ACDEB. After this we have ACDEF.
So, rank of ACDEF = 4718

The question is "what would be the rank of “ACDEF’?"

Hence the answer is "4718"

Choice C is the correct answer

This is a very popular template. This question is based on counting the number of times a particular digit appears in a list.

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 units 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 units 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 question is "If we listed all numbers from 100 to 10,000, how many times would the digit 3 be printed?"

Hence the answer is "3980"

ATTITUDE has 8 letters, of which 3 are Ts
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 Ts 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 Ts 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.

The question is "In how many ways can letters the word ATTITUDE be rearranged such that no two Ts are adjacent to each other?"

Hence the answer is "2400"

Choice B is the correct answer.

Let us find the one pair of values for a, b.
a = 4, b = 19 satisfies this equation.
2*4 + 5*19 = 103.

Now, if we increase ‘a’ by 5 and decrease ‘b’ by 2 we should get the next set of numbers. We can keep repeating this to get all values.
Let us think about why we increase ‘a’ by 5 and decrease b by 2.
a = 4, b = 19 works.

Let us say, we increase ‘a’ by n, then the increase would be 2n.
This has to be offset by a corresponding decrease in b.
Let us say we decrease b by ‘m’.
This would result in a net drop of 5m.
In order for the total to be same, 2n should be equal to 5m.
The smallest value of m, n for this to work would be 2, 5.

a = 4, b = 19
a = 9, b = 17
a = 14, b = 15
..
And so on till
a = 49, b = 1
We are also told that ‘a’ should be greater than ‘b’, then we have all combinations from (19, 13) … (49, 1).
7 pairs totally.

The question is "2a + 5b = 103. How many pairs of positive integer values can a, b take such that a > b?"

Hence the answer is "7"

If we want to have two throws to have the same outcome and the other two being something different, we are essentially looking for three different outcomes in each throw.
We can have three different outcomes in each throw in ⁶C₃ ways. We can select the one that is being repeated out of the three outcomes in ³C₁ ways.

And then, they can be arranged in 4!/2! ways, since there are totally 4 throws out of which 2 are repeated. So, Totally , we can do this in ⁶C₃ * ³C₁ * 4!2!4!2! = (4∗5∗6)6(4∗5∗6)6 * 3 *12 = 20*3*12 = 720 ways.

The question is "In how many outcomes do we have two throws have the same number and the other two something different?"

Hence the answer is "720"

Choice A is the correct answer.