Algebra Question and Solution

Question

|x| + |2y| + |3z| = 13, x * y * z is non-zero; x, y, z are all integers. How many sets of values are possible?

Answer
64 sets of values

Explanation

CAT Counting: Letters and Words

Question

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

A.      4716

B.      4720

C.      4718

D.      4717

Correct Answer

Choice C

Explanatory Answer

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

Answer Choice (C)         

CAT Counting: Question with digits

This is an interesting question from Counting. Simple framework, but one needs to be very careful with the enumeration. One can get wrong answers in a number of ways.

Question

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

A.      3980

B.      3700

C.      3840

D.      3780

Correct Answer

Choice A

Explanatory Answer

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 or10’s place or units place.

Ø  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

Ø  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

Ø  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 or10’s place or units place.

Ø  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.

Ø  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.

Ø  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.

Ø  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

Answer Choice (A)

Permutation and Combinations - Theory

The previous two posts discussed a key idea behind counting number of options for the question of type a + b +  c = r. Now, let us  see a general version of this.

Key formulae
For a1 + a2 + a3 ....ar = n. The number of solutions where a1,a2,a3....ar all take natural numbers is (n-1) C (r-1).

For a1 + a2 + a3 ....ar = n. The number of solutions where a1,a2,a3....ar can take all whole number values is  (n+r-1) C (r-1).


Similar questions on the same theme
The proofs are just an extension of the idea discussed in the two posts. Now, this kind of question can get asked in different frameworks. Few examples are given below.

1. 10 identical toys need to be placed in 3 distinct boxes. In how many ways can this be done?
2. Mark needs to pick up exactly 40 fruits for his family. He has to pick at least 8 Apples, at least 6 Oranges and at least 5 mangoes. He should not pick up any other fruit. In how many ways can this be done?

Permutations and Combinations

Question
Sum of three whole numbers a, b and c is 10. How many ordered triplets (a, b, c) exist? (This is a variant on the previous question that can be found here). Will post the solution in a day or two.


Correct Answer
12C2

Explanatory Answer

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. The discussion can be seen here .

We cannot follow a similar approach as a,b,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

|     |     |     |    |     |     |     |     |     |     |     |     |

This question now becomes the equivalent of placing two '+' symbols somewhere between these sticks. For instance

|     |     |     | + |     |     |     |     |  + |     |     |     |, 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.

Questions and solutions on Permutation and Combination

Permutation and Combination.

Questions
1. Sum of three natural numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?

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

Correct Answers:
1. 9C2
2. 10C2


Explanatory Answers

1. Sum of three natural numbers a, b and c is 10. How many ordered triplets (a, b, c) exist?

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

|     |     |     |    |     |     |     |     |     |

This question now becomes the equivalent of placing two '+' symbols somewhere between these sticks. For instance

|     |     |     | + |     |     |     |     |  + |, 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.


2. In how many ways 11 identical toys be placed in 3 distinct boxes such that no box is empty?
This is nothing but number of ways of having a,b,c such that a + b + c = 11, where a, b, c are natural numbers.

10C2. 

The two questions stated above are almost identical ones.

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