CAT 2025 Lesson : Permutations & Combinations - Using Blanks

A new flag is designed with six vertical stripes using some or all of the colours – yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is
[CAT 2004]

(1) $12 \times 81$ (2) $16 \times 192$ (3) $20 \times 125$ (4) $24 \times 216$

Solution

Below is a representation of the flag. There are a total of $4$ colours available. The general condition is that adjacent stripes should not be of the same colour. The stripes have been named from A to F, for use in explanation only. And, below each letter is the number of ways in which that stripe can be coloured.

When we start from stripe A, we can use any of the $\bold{4}$ colours in stripe A.
The next stripe B cannot have A's colour. So, $\bold{3}$ colours can be used in stripe B.
The next stripe C cannot have B's colour. So, $\bold{3}$ colours can be used in stripe C.
The same logic applies for D, E and F stripes, which can take $3$ colours each.

A	B	C	D	E	F
4	3	3	3	3	3

Going forward, we will use blanks to explain the same, wherein the position of each blank will correspond to each of the stripes from A to F. In these blanks, we will fill the the number of ways that blank (in this case stripe) can be filled. Product of these blanks is the total ways in which the flag can be coloured

In this question, number of ways $= \underline{4} \times \underline{3} \times \underline{3} \times \underline{3} \times \underline{3} \times \underline{3} = 12 \times 81$

Answer: ($1$) $12 \times 81$

How many $4$-digit numbers exist such that
(I) no $2$ digits of the number are equal?
(II) the number remains the same when the digits are reversed?
(III) they are less than $6000$ and multiples of $5$?
(IV) they are multiples of $4$ and none of their digits are $4$, $5$, $6$, $7$ or $8$?

Solution

Case I
The digits of the number are distinct and cannot be repeated. In a $4$-digit number, the left-most blank or thousands digit cannot be zero. This is a blank with a condition.

∴ Thousands place has to be filled first and can be filled with any of the $\bold{9}$ digits (digits from $1$ to $9$).

After filling the thousands place, we are now left with $\bold{9}$ digits to fill the hundreds place, including 0.
Similarly, we are left with $\bold{8}$ digits to fill the tens place and $\bold{7}$ digits to fill the units place.

$\therefore$ Total Permutations $= \underline{9} \times \underline{9} \times \underline{8} \times \underline{7} = 4536$

Case II
If the 4-digit number remains the same when digits are reversed, then the thousands and units digits are the same, and the hundreds and tens digits are the same. Note that digits can be repeated in this question.

Thousands place can be filled in $\bold{9}$ ways (digits from $1$ to $9$) and hundreds place in $10$ ways ($0$ to $9$). Tens and Units places can be filled in $\bold{1}$ way each, i.e. the same as the hundreds and thousands respectively.

$\therefore$ Total Permutations $= \underline{9} \times \underline{10} \times \underline{1} \times \underline{1} = 90$

Case III
Thousands place can only take $5$ values($1,2,3,4$ and $5$), hundreds and tens place can take all the $10$ digits and the units place can take only $2$ values ($0$ and $5$).

$\therefore$ Total Permutations $= \underline{5} \times \underline{10} \times \underline{10} \times \underline{2} = 1000$

Case IV
Only the digits $0, 1, 2, 3, 9$ can be used. For a number to be divisible by $4$, the last two digits should be divisible by $4$. Therefore, the last $\bold{2}$ digits (tens and units places) can only take $\bold{5}$ values - $00, 20, 12, 32, 92$. As repetition is allowed, thousands place can take $\bold{4}$ values ($1, 2, 3, 9$) and hundreds place can take $\bold{5}$ values.

$\therefore$ Total Permutations $= \underline{4} \times \underline{5} \times \underline{5} = 100$

Answer: (I) $4536$; (II) $90$; (III) $1000$; (IV) $100$

When $3$ dice are rolled, in how many ways can we have a prime number on the first die, even number on the second and a multiple of $3$ on the third die?

Solution

The first die can take $3$ digits ($2, 3, 5$), the second die can take $3$ digits ($2, 4$ and $6$) and the third die can take two digits ($3$ and $6$).

$\therefore$ Total Permutations $= \underline{3} \times \underline{3} \times \underline{2} = 18$

Answer: $18$

Hanuram has $3$ bats $-$ cricket bat, baseball bat, softball bat, and $3$ balls $-$ cricket ball, baseball and softball.
(I) In how many different ways can Hanuram choose a bat and a ball from this set?
(II) If Hanuram is allowed to choose only $1$ item, then in how many ways can he choose a bat or a ball?

Solution

Case I: Hanuram has to choose $1$ bat and $1$ ball. As there are $3$ bats and $3$ balls, there are $3$ ways in which he can choose a bat and $3$ ways in which he can choose a ball. Using blanks, the number of ways of
choosing a bat and a ball $= \underline{3} \times \underline{3} = 9$

Case II: In this case also Hanuram has $3$ choices for bats and three choices for balls. However, he needs to pick only $1$ item. As there are a total of $3 + 3 = 6$ items, he can choose in $6$ ways.

Answer: (I) $9$; (II) $6$

Maran has $3$ formal shirts, $4$ formal trousers, $5$ t-shirts, $6$ jeans, $2$ pairs of formal shoes and $3$ pairs of sandals. He can either dress up all formal $-$ wearing a formal shirt with formal trousers and a pair of formal shoes, or dress up casual $-$ wearing a t-shirt with jeans and a pair of sandals. In how many ways can Maran get dressed for the party?

Solution

Maran can dress formally or casually for the party. So, we need to add the number of ways he can dress formally with those of casually. To dress formally, he has to wear a formal shirt and formal trousers and a pair of formal shoes. So, we multiply the number of different items of each type he has.
Number of ways of dressing up formally for the party $= \underline{3} \times \underline{4} \times \underline{2} = 24$

Likewise to dress casually, he has to wear a t-shirt and jeans trousers and a pair of sandals.
Number of ways of dressing up casually for the party $= \underline{5} \times \underline{6} \times \underline{3} = 90$

Total number of ways in which he can dress formally or casually $= 24 + 90 = 114$

Answer: $114$

How many $4$-digit even numbers exist such that no two digits are the same?

Solution

To solve these questions, we first need to understand where the condition is and begin by filling the blanks with the conditions. The condition lies in the thousands digit (which cannot have $0$, as it would make it a $3$-digit number) and units digit (which needs to be even). So, we can have $2$ approaches here. Approach 1 starts with the thousands digit and Approach 2 starts with the units digit.

Approach 1: When we begin with the thousands place, what goes into this impacts the number of ways in which you can fill the units place. Therefore, we break the thousands place cases on the basis of the condition governing the units place. Note that no repetition of digits is allowed.

Cases where thousands place is even
Thousands can be filled by either of the non-zero even digits ($2, 4, 6, 8$). After this, we are left with $4$ even digits (including $0$) to fill the units place. We can then fill the hundreds place with either of the $8$ remaining digits, which leaves us with $7$ digits for the tens place. Number of Permutations $= \underline{4} \times \underline{8} \times \underline{7} \times \underline{4} = 896$

Cases where the thousands place is odd
Thousands place can take any of the $5$ odd digits ($1, 3, 5, 7, 9$). Units place can take any of the $5$ even digits ($2, 4, 6, 8, 0$). The hundreds and tens places can take $8$ and $7$ digits respectively. Number of Permutations $= \underline{5} \times \underline{8} \times \underline{7} \times \underline{5} = 1400$

Total Permutations $= 896 + 1400 = \bold{2296}$

Approach 2: The thousands place cannot be $0$. Here, we start with filling the units digit and break into two cases – one where the units place is $0$ and the other where it is not.

Cases where Units place is 0
$1$ way to fill the units place. The thousands, hundreds and tens can be filled in $9, \space 8$ and $7$ ways respectively.
Number of Permutations $= \underline{9} \times \underline{8} \times \underline{7} \times \underline{1} = 504$

Cases where Units is place is not 0
$4$ ways to fill units place ($2, 4, 6, 8$). Thousands place can be filled in $8$ ways (digits other than $0$ and that occupying the units place). The hundreds and tens places can take $8$ and $7$ digits respectively. Number of Permutations $= \underline{8} \times \underline{ 8} \times \underline{7} \times \underline{4} = 1792$

Total Permutations $= 504 + 1792 = \bold{2296}$

Answer: $2296$