Quiz Level-1

Hint

Number of ways in which the first book can be placed = 7

Number of ways in which the second book can be placed = 6

Similarly,

The total number of ways in which seven books can be arranged on a shelf = 7 × 6 × 5 × 4 × 3 × 2 × 1 (i.e., 7!)

= 5040

Hint

Given word: MATHEMATICS

Number of letters = 11

M = 2

A = 2

T = 2

Number of different arrangements = 11!/(2! 2! 2!)

= (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(2 × 1 × 2 × 1 × 2 × 1)

= 4989600

Hint

Given digits: \(1,2,3,4,6,7\)
Number of digits \(=6\)
Number of possible digits at unit’s place \(=3(2,4\) and 6\()\)
\(\Rightarrow\) Number of permutations \(={ }^3 \mathrm{P}_1=3\)
When one of the digits is taken in units’ place, then the number of possible digits available \(=5\)
\(\Rightarrow\) Number of permutations \(={ }^5 P_2=5 ! /(5-2) !=5 ! / 3 !=120 / 6=20\)
The total number of permutations \(=3 \times 20=60\).
Therefore, 60 three-digit numbers can be made using the given digits.

Hint

8 people have equal chances to get 1st prize gold, 7 people are having equal chances of getting the 2nd prize silver, and 6 people are having equal chances of getting 3rd prize.

Hence the total number of ways  =  8 ⋅ 7 ⋅ 6   =  336 ways

Hint

1^st man can wear any of the 4 coats.
2^nd man can wear any of the remaining 3 coats.
3^rd man can wear any of the remaining 2 coats.

So, number of ways in which 3 men can wear 4 coats 

  =  4 ⋅ 3⋅ 2 

  =  24

Similarly,

Number of ways in which 3 men can wear 5 waistcoats 

  =  5 ⋅ 4 ⋅ 3

  =  60

Number of ways in which 3 men can wear 6 caps

  =  6 ⋅ 5 ⋅ 4

  =  120

Hence, required number of ways

  =  24 ⋅ 60 ⋅ 120

  =  172800

Hence the total number of ways is 172800.

Hint

Number of letters in the word “SIMPLE” \(=6\)
All are unique letters.
Number of permutation \(={ }^6 p_6=6\) !
\(
\begin{gathered}
=6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \\
=720
\end{gathered}
\)
Hence the total number of permutations is 720.

Hint

Each question has four choices.
Number of ways to answer \(1^{\text {st }}\) question \(=4\)
Number of ways to answer \(2^{\text {nd }}\) question \(=4\)
Number of ways to answer \(3^{\text {rd }}\) question \(=4\)
and son on ……
Number of ways to answer \(10^{\text {th }}\) question \(=4\)

\(
\begin{gathered}
\text { Number of ways }=4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4 \\
=4^{10}
\end{gathered}
\)
Hence the total number of ways \(=4^{10}\)

Hint

The first four questions have three choices and the remaining have five choices.
Number of ways to answer \(1^{\text {st }}\) question \(=3\)
Number of ways to answer \(2^{\text {nd }}\) question \(=3\)
Number of ways to answer \(3^{\text {rd }}\) question \(=3\)
Number of ways to answer \(4^{\text {th }}\) question \(=3\)
Number of ways to answer \(5^{\text {th }}\) question \(=5\)
Number of ways to answer \(6^{\text {th }}\) question \(=5\)
Number of ways \(=3 \cdot 3 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5\)
\(=3^4 \cdot 5^6\)
Hence the total number of ways is \(3^4 \cdot 5^6\).

Hint

Question number \(n\) has \(n+1\) choices.
Number of ways to answer \(1^{\text {st }}\) question \(=2\)
Number of ways to answer \(2^{\text {nd }}\) question \(=3\)
Number of ways to answer \(3^{\text {rd }}\) question \(=4\)
Number of ways to answer \(4^{\text {th }}\) question \(=5\)

and so on………..
Number of ways to answer \(10^{\text {th }}\) question \(=11\)
Number of ways \(=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \cdot 11\)
\(=11 !\)
Hence the total number of ways is \(11 !\).

Hint

The given word contains 5 different letters.
Required number of ways \(={ }^5 P_5 = 5 !=(5 \times 4 \times 3 \times 2 \times 1)=120\).

Hint

The word ‘PRESENT’ contains 7 letters, namely \(2 \mathrm{E}\) and all other 5 are different.
\(
\therefore \quad \text { Required number of ways }=\frac{7 !}{{2 !}}=\frac{7 \times 6 \times 5 \times 4 \times 3 \times({2!})}{{2!}}=2520 \text {. }
\)

Hint

The word ‘ENGINEERING’ contains 11 letters, namely \(3 \mathrm{E}, 3 \mathrm{~N}, 2 \mathrm{G}, 2 \mathrm{I}\) and \(1 \mathrm{R}\).
\(
\begin{aligned}
& \therefore \quad \text { Required number of ways }=\frac{11 !}{3 ! \times 3 ! \times 2 ! \times 2 ! \times 1 !} \\
& \quad=\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 6 \times 2 \times 2 \times 1} \\
& \quad=(11 \times 10 \times 7 \times 6 \times 5 \times 4 \times 3)=277200
\end{aligned}
\)

Hint

The given word ‘DESIGN’ contains 4 consonants and 2 vowels.
At the two ends the two vowels can be arranged in 2 ways.
The remaining 4 letters can be arranged in 4! = (4 × 3 × 2 × 1) = 24 ways.
Total number of ways = (24 × 2) = 48.
∴ The required number of ways = 48.

Hint

The given word contains 8 different letters.
When the vowels AUE are taken together, we may treat them as 1 letter.
Then, the letters to be arranged are DGHTR (AUE)
The vowels can be arranged in \({ }^6 P_6=6!=720\) ways.
The vowels AUE may be arranged in \(3!=6\) ways.
Required number of ways \(=(720 \times 6)=4320\) ways.

Hint

Well, there are 10 possible digits for each digit in the code, and the code is four digits. So there are ${\mathrm{10\; x\; 10\; x\; 10\; x\; 10}}^{}$ possible PIN codes.

Hint

so Required number of ways=
\(
\begin{aligned}
& { }^n P_r=n ! /(n-r) ! \\
& { }^7 P_3=7 ! /(7-3) !=4 ! \cdot 5 \cdot 6 \cdot 7 / 4 !=210
\end{aligned}
\)

Hint

Here \(n=10, r=4\)
so required number of ways= \({ }^n P_r=n ! /(n-r) !\)
\(
{ }^{10} p_4=10 ! /(10-4) !=6 ! 7 \cdot 8 \cdot 9 \cdot 10 / 6 !=5040
\)
In 5040 ways 4 women can be chosen as team leaders.

Hint

Here \(n=26, r=4\)
so required number of ways =
\(
{ }^n P_r=n ! /(n-r) !
\)
\(
{ }^{26} p_4=26 ! /(26-4) !=22 ! .23 \cdot 24 \cdot 25 \cdot 26 / 22 !=358800
\)
In 358800 ways, 4 different letter permutations are available.

Hint

Here \(n=10, r=3\)
so Required number of ways=
\(
\begin{aligned}
& { }^n P_r=n ! /(n-r) ! \\
& { }^{10} P_3=10 ! /(10-3) !=7 ! \cdot 8 \cdot 9 \cdot 10 / 7 !=720
\end{aligned}
\)
In 720 ways, three-digit permutations are available.

Hint

Here \(n=18, r=3\)
so Required number of ways=
\(
\begin{aligned}
& { }^n P_r=n ! /(n-r) ! \\
& { }^{18} P_3=18 ! /(18-3) !=15 ! \cdot 16 \cdot 17 \cdot 18 / 15 !=4896
\end{aligned}
\)
Among the 18 contestants, in 4896 number of ways, a judge can award a 1st, 2nd and 3rd place in a contest.

Hint

Here \(n=7, r=7\)
so required number of ways=
\(
\begin{aligned}
& { }^n P_r=n ! /(n-r) ! \\
& { }^7 P_7=7 ! /(7-7) !=7 ! / 0 !=5040
\end{aligned}
\)
In 5040 number of ways, 7 people can organize themselves in a row.

Hint

Total number of contestants \(=12\)
To find \(P(12,3)\) so as to calculate the total count of possible outcomes for the top 3.
\(
\begin{aligned}
& P(12,3)=12 ! /(12-3) ! \\
& =1,320 \text { outcomes }
\end{aligned}
\)

Hint

To calculate the ordered subset of five players ( \(r)\) from the given set of ten players \((n)\).
\(
\begin{aligned}
& P(10,5)=10 ! /(10-5) ! \\
& =30,240 \text { orders }
\end{aligned}
\)

Hint

Well, one possible assignment of the diets to the pigs would be for the first five pigs to be placed on diet A, the second five pigs to be placed on diet B, and the last 5 pigs to be placed on diet C. That is:
A A A A A B B B B B C C C C C
Another possible assignment might look like this:
A B C A B C A B C A B C A B C
Upon studying these possible assignments, we see that we need to count the number of distinguishable permutations of 15 objects of which 5 are of type A, 5 are of type B, and 5 are of type C. Using the formula, we see that there are:
\(
\frac{15 !}{5 ! 5 ! 5 !}=756756
\)
ways in which 15 pigs can be assigned to the 3 diets.

Hint

There are nine total letters in word, therefore, \(n=9\)
\(
\begin{aligned}
& n_1=1 \mathrm{T} \\
& n_2=4 \mathrm{E} \\
& n_3=2 \mathrm{N} \\
& n_4=2 \mathrm{S}
\end{aligned}
\)
Therefore, number of distinguishable permutations equals:
\(
\begin{aligned}
& \frac{n !}{n_{1} ! n_{2} ! \ldots n_{k} !} \\
& =\frac{9 !}{1 ! 4 ! 2 ! 2 !} \\
& =\frac{(9)(8)(7)(6)(5)}{4} \\
& =3,780
\end{aligned}
\)

Hint

There are nine total letters in word, therefore, \(n=10\)
\(
\begin{aligned}
& n_1=2 \quad{ }\mathrm{C} \\
& n_2=3 \quad \mathrm{T} \\
& n_3=3 \quad \mathrm{N} \\
& n_4=1 \quad \text {A} \\
& n_5=1 \quad \mathrm{T}
\end{aligned}
\)
Therefore, number of distinguishable permutations equals:
\(
\begin{aligned}
& \frac{n !}{n_{1} ! n_{2} ! \ldots n_{k} !} \\
& =\frac{10 !}{2 ! 3 ! 3 ! 1 ! 1 !} \\
& =\frac{(10)(9)(8)(7)(6)(5)(4)}{(2)(1) \cdot(3)(2)(1)} \\
& =50,400
\end{aligned}
\)

Hint

Our set, in this case, is the committee. The size of this set is 15 people. So n =15. The size of our permutation will be 3 people since we want to create a 3-person board. The reason why we know this is a permutation is that each person is being assigned a specific position. For example, let’s say our three people chosen are Lisa, Michael, and Faith. There are 6 different possible ways that we can assign the positions to these three: there are 3 choices for secretary, then only 2 choices for treasurer, and the last person will be made chairperson. Since $n$ and r=, then the total number of boards that can be created is $=15\times 14\times 13=2730.$

Hint

P(12, 3) = 12 × 11 × 10 = 1, 320.

Hint

\(
\mathbf{P}(15,3)=15 \cdot 14 \cdot 13=2,730 \text {. }
\)

Hint

Thus the number of different words we can form by rearranging the letters must be
\(
=\frac{4 !}{2 !} = 12
\)
Note that 2 ! counts the number of ways we can permute the two E’s in any given arrangement.

Hint

\(\frac{6 !}{3 !}=\frac{720}{6}=120\). There are 6 letters in ALPACA and one of them, ‘A’ is repeated 3 times.

Hint

\(
\frac{10 !}{1 ! \cdot 3 ! \cdot 2 ! \cdot 2 ! \cdot 1 ! \cdot 1 !}=151,200 \text {. }
\)
There are 10 letters in BOOKKEEPER. In alphabetical order, \(\mathrm{B} \leftrightarrow 1, \mathrm{E} \leftrightarrow 3, \mathrm{~K} \leftrightarrow 2, \mathrm{O} \leftrightarrow 2, \mathrm{P} \leftrightarrow 1, \mathrm{R} \leftrightarrow 1\).
Note that the total number of letters is the sum of the multiplicities of the distinct letters: \(10=1+3+2+2+1+1\).

Hint

The number of objects, here is 5 , because the word SMOKE has 5 alphabets.
Also, \(\mathrm{r}=3\), as 3 letter-word has to be chosen.
Thus the permutation will be
\(=n^r\) as repetition is allowed.
\(
=5^3=125
\)

Hint

In these types of questions, we assume all the vowels to be a single character, i.e., “IE” is a single character. So, now we have 5 characters in the word, namely, D, R, V, R, and IE. But, R occurs 2 times.
\(\Rightarrow>\) Number of possible arrangements \(=5 ! / 2 !=60\) Now, the two vowels can be arranged in \(2 !=2\) ways.
=> Total number of possible words such that the vowels are always together \(=60 \times 2=120\)

Hint

We just have to use the permutations formula and replace the values \(n=10\) y \(r=4\) :
\(
\begin{gathered}
{ }^n P_r=\frac{n !}{(n-r) !} \\
=\frac{10 !}{(10-4) !} \\
=\frac{10 !}{(6) !} \\
=\frac{10 \times 9 \times 8 \times 7 \times 6 !}{(6) !}=5040
\end{gathered}
\)