Quiz

Hint

\(
\text { The value of } 7 \text { ! is } 5040 \text {, i.e. } 7 \text { ! }=7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=5040 \text {. }
\)

Hint

\(
3 ! =3 \cdot 2 \cdot 1=6
\)

Hint

\(
10 !=3628800
\)

Hint

The given word contains 7 letters, all different. \(\therefore\) Required number of ways \(=7 !\)
\(
=(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)=5040 \text {. }
\)

Hint

The given word contains 7 letters of which \(D\) is taken 3 times. \(\therefore\) Required number of ways \(=\frac{7 !}{3 !}=\frac{7 \times 6 \times 5 \times 4 \times {3 !}}{3 !}\) \(\quad=(7 \times 6 \times 5 \times 4)=840\).

Hint

The given word contains 6 letters of which E is taken 2 times.

\(
\frac{6 !}{{2 !}}=\frac{6 \times 5 \times 4 \times 3 \times {2 !}}{2 !}=360 .
\)

Hint

The given word contains 5 different letters.
Keeping the vowels UE together, we suppose them as 1 letter.
Then, we have to arrange the letters JDG(UE).
Now, 4 letters can be arranged in \({4 !}=24\) ways.
The vowels (UE) can be arranged among themselves in 2 ways.
\(\therefore\) Required number of ways \(=(24 \times 2)=48\).

Hint

The given word contains 7 different letters. Keeping the vowels (AUIO) together, we take them as 1 letter.
Then, we have to arrange the letters CTN(AUIO).
\(\text { Now, } 4 \text { letters can be arranged in } {4 !}=24 \text { ways. }\)
The vowels (AUIO) can be arranged among themselves in 4 ! = 24 ways.
∴ Required number of ways = (24 × 24) = 576.

Hint

Taking the vowels (EA) as one letter, the given word has the letters XTR (EA), i.e., 4 letters.
These letters can be arranged in \({4 !}=24\) ways.
The letters \(E A\) may be arranged amongst themselves in 2 ways.
The number of arrangements having vowels together \(=(24 \times 2)=48\) ways.
Total arrangements of all letters \(={5 !}=(5 \times 4 \times 3 \times 2 \times 1)=120\).
Number of arrangements not having vowels together \(=(120-48)=72\)

Hint

Number of ways of attempting 1st, 2nd, \(3^{\text {rd }}\) questions are each. Total number of ways \(=4^3=4 \times 4 \times 4=64\)
Number of ways, getting all correct answers \(=1^3=1\)
\(\therefore\) Number of ways of not getting all answers correct
\(=64-1=63\)

Hint

There are 2 primary teachers.
They can stand in a row in = 2! = 2 × 1 ways = 2 ways
Two middle teachers.
They can stand in a row in = 2! = 2 × 1 = 2 ways
There are two secondary teachers.
They can stand in a row in = 2! = 2 × 1 = 2 ways
These three sets can be arranged in 3! ways = 3 × 2 × 1 = 6 ways
Hence, the required number of ways = 2 × 2 × 2 × 6 = 48 ways

Hint

Number of persons available to sit in the first chair = 15

Number of persons available to sit in the second chair = 14

It keeps going in this way,

So, the total number of ways to sit 15 students = 15!

Hint

Number of greyhounds will take the 1st position = 8
The number of greyhounds will take the 2nd position = 7
Continuing in this way, we get
8 ! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= 40320

Hint

Total number of candidates = 10
To choose the first strongest candidate, there are 10 options. To choose the second strongest candidate, there are 9 options available. To choose the third strongest candidate, there are 8 options available. To choose the fourth strongest candidate, there are 7 options available, So, we get
= 10 x 9 x 8 x 7
= 5040
So, the total number of ways to select 4 candidates, we have 5040 ways.

Hint

Total number of horses = 8
Number of persons who are having the chance to reach the first position = 8
Number of persons who are having the chance to reach the second position = 7
Number of persons who are having the chance to reach the third position = 6
= 8 x 7 x 6
= 336
So, the total number of ways is 336.

Hint

Given set = {a, b, c, d, e, f, g, h}
Total number of distinct characters = 8
The number of options to fix the 1st letter of the password = 8
The number of options to fix the 2nd letter of the password = 7
The number of options to fix the 3rd letter of the password = 6
The number of options to fix the 4th letter of the password = 5
The number of options to fix the 3rd letter of the password = 4
= 8 x 7 x 6 x 5 x 4
= 6720
So, the total number of ways = 6720.

Hint

Number of greyhounds = 6
Number of greyhounds to fill the 1st position = 6
Number of greyhounds to fill the 2nd position = 5
Number of greyhounds to fill the 3rd position = 4
= 6 x 5 x 4
= 120

Hint

Number of persons = 8
Number of persons who occupies the position of Chairperson = 8
Number of persons who occupies the position of Secretary = 7
Number of persons who occupies the position of Treasurer = 6
The total number of outcomes = 8 x 7 x 6 = 336

Hint

Total number of letters in the given word = 6
Excluding two A’s, we have B, C, U and S. We consider 4 + 1 = 5 units.
Total number of ways = 5!
= 5 x 4 x 3 x 2 x 1
= 120

Hint

Total number of letters in the given word \(=7\)
To arrange those 7 letters, we have 7 ! ways. Here 2 terms cannot be separated.
\(
\begin{gathered}
\text { Total number of ways }=7 ! / 2 ! \\
=(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) /(2 \times 1) \\
=2520
\end{gathered}
\)

Hint

Given word : REMAND
Number of letters in the word \(=6\)

a) Without restrictions:
Number of ways, that letters can be arranged \(=6\) !

b) they begin with RE?
\(
R E _ _ _ _
\)
Number of words to be arranged \(=4\)
Number of ways available \(=4\) !

c) they do not begin with RE?
Number of words can be created, that doesn’t start with RE
\(
\begin{gathered}
=6 !-4 ! \\
=4 !(6 \times 5-1) \\
=4 ! \times 29 = 696
\end{gathered}
\)

Hint

a) Without restriction :
Number of boys = 5, number of girls = 4
Total number of members = 5 + 4
= 9
Without restrictions, we have = 9! ways.

b) Boys and girls to be seated in alternate positions.
= 5! x 4! ways

Hint

a) Number of available options = 7 x 6 x 5 x 4 x 3 x 2 x 1
= 5040

b) Number of ways to choose 1st, 2nd and 3rd position :
= 7 x 6 x 5
= 210

Hint

\(
n=2 \cdot 5 !=240
\)

Explanation: Let the 6 persons be \(A, B, C, D, E\), and \(F\).
If two persons ( \(A\) and \(C\)) sitting beside each other, then let’s club them into one unit, K
We have
\(B, D, E, F, K(\) total 5)
Now we have 5 ! ways = 120
Now \(\mathrm{K}\) can be arranged in 2! ways.
Number of ways
\(
\begin{aligned}
& =5 ! \times 2 ! \\
& =120 \times 2 \\
& =240
\end{aligned}
\)

Hint

\(
n=2 \cdot 3 !=12
\)

Explanation: The two-person seating together consider as 1, so there are total 3! = 6 possible ways. The two-person seating together there are 2! ways. Therefore the total number of ways to seat = 3! x 2! = 12 ways.

Hint

Total possible number of words \(4!=4 \cdot 3 \cdot 2 \cdot 1=24\)

Hint

As one position already fixed, we have \(n=(5-1) !=24\) ways.

Hint

\(
n=(8-2) \cdot(8-2-1) \cdot(8-2-2) \cdot(8-2-3)=360
\)

Hint

\(
n=4 \cdot 3 \cdot 2 \cdot 1=24
\)

Hint

\(n=5 !=120\)

Hint

\(
n=7 !=5040
\)

Hint

\(
\begin{aligned}
& x !: 5=1008 \\
& x !=5040 \\
& x=7
\end{aligned}
\)

Hint

\(n=10 !=3628800\)

Hint

\(n=\frac{4 !}{2 !}=12\)

Hint

\(
\begin{aligned}
& 2^3=8 \\
& 8 / 4=2 \\
& x=2
\end{aligned}
\)

Hint

\(
n=4 !=24
\)

Hint

\(
n=\frac{(4+5+6) !}{4 ! \cdot 5 ! \cdot 6 !}=630630
\)

Hint

\(
n=6 \cdot 5 \cdot 4=120
\)

Hint

\(
\begin{aligned}
& n=6 \\
& x=n !=6 !=720
\end{aligned}
\)

Hint

\(
n=2 \cdot 1=2
\)

Hint

\(
n=\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{3 !}=1120
\)

Hint

\(
n=6 !=720
\)

Hint

\(
n=10 !=3628800
\)

Hint

\(
n=\frac{(2+2) !}{2 ! \cdot 2 !}=6
\)

Hint

\(
n=2 \cdot(7-1) !=1440
\)

Note: Consider 1 and 2 as one having 2 ways and the rest 6!(7-1)!.

Hint

\(
n=\frac{19 !}{2 ! \cdot 2 ! \cdot 2 ! \cdot 3 ! \cdot 3 ! \cdot 3 ! \cdot 4 !}=2933186256000
\)

Hint

\(
n=16 \cdot 15 \cdot 14=3360
\)

Hint

\(
n=3 \cdot(8+3-2) ! \cdot(3-1)=2177280
\)

Hint

Hint

\(
n=2 \cdot 6 !=1440
\)

Hint

\(
n=\frac{5 !}{3 ! \cdot 2 !}=10
\)

Hint

\(
n=(11-4) \cdot(11-4-1) \cdot(11-4-2) \cdot(11-4-3) \cdot(11-4) / 5 !=49
\)

Hint

\(
n=\frac{(3+2+1) !}{3 ! \cdot 2 ! \cdot 1 !}=60
\)

Hint

\(
n=4 \cdot 3 \cdot 2=24
\)

Hint

\(
n=1 \cdot 4 \cdot 4 \cdot 3 \cdot 2 \cdot 1=96
\)

Hint

\(
n=9 \cdot 8 \cdot 7=504
\)

Hint

\(
n=(4 \cdot 3 \cdot 2 \cdot 1) \cdot(3 \cdot 2 \cdot 1)=144
\)

Hint

\(
n=4 !=24
\)

Hint

\(
n=\frac{4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1}=12
\)

Hint

\(
n=5 \cdot 4 !=120
\)

Hint

\(
\begin{aligned}
& O I A \\
& \text { PTCL } \\
& n_1=(4+1) !=120 \\
& n_2=3 !=6 \\
& n=n_1 \cdot n_2=120 \cdot 6=720
\end{aligned}
\)

Explanation:

the word OPTICAL has three vowels and 4 consonants. Since all the three vowels are to be together, mark as one symbol X This with the 4 consonants become 5 which can be permuted in \(5 !=120\) ways. The three vowels in \(\mathrm{X}\) can be permuted among themselves in \(3 !=6\) ways.
The total number of arrangements of the letters of the word OPTICAL so that all the vowels are always together is \(120 \times 6=720\).

Hint

\(
n=4 \cdot 3 \cdot 2 \cdot 1=24
\)

Hint

\(
\begin{aligned}
& n=4 \\
& x=n(n-1)(n-2) \\
& x=4 \cdot 3 \cdot 2=24
\end{aligned}
\)

Hint

\(
\begin{aligned}
& a=6 \\
& b=6 \\
& c=5 \\
& n_1=a !=6 !=720 \\
& n_2=b !=6 !=720 \\
& n_3=c !=5 !=120 \\
& n=3 ! \cdot n_1 \cdot n_2 \cdot n_3=3 ! \cdot 720 \cdot 720 \cdot 120=373248000=3.732 \cdot 10^8
\end{aligned}
\)

Explanation:

Andrea, first, calculate how many ways to arrange each type of book individually. Each algebra book has 6! ways of being arranged. Each geometry book has 6! ways of being arranged. Each statistics book has 5! ways of being arranged. Now, there are three groups of books, so there are 3! ways of arranging them on the shelf. 6! x 6! x 5! x 3! = 373248000

Hint

This problem is identical to saying if you have 7 boys in a row, how many ways are there to place 5 girls in that row without having two girls next to each other? you would have 8 locations to choose from.

B B B B B B B
+ B + B + B + B + B + B + B +

Here you have 8 places to arrange the girls and 5 girls. This is the same as 8 choose 5 which equals 56.

Then for each of those 56 arrangements, you can look at every possible arrangement of individual girls, the number of which is 5!

Lastly for each of those unique placements of girls, and unique orders of girls, you can take the unique order of boys which would be 7!

In my calculation, the total should be 56 x 5! x 7! = 33868800

Hint

\(
n=10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5=151200
\)

Hint

BRATISLAVA
Number of A: 3
\(
n=\frac{10 !}{3 !}=604800
\)

Hint

\(
n=\frac{6 !}{3 ! \cdot 2 !}=60
\)

Hint

\(
n=\frac{6 !}{2 ! \cdot 4 !}=15
\)

Hint

\(
n=4 !=24
\)