Permutations Combinations 1-1

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Permutations and Combinations: FormulasPermutations and Combinations: Exercise - 1Permutations and Combinations: Exercise - 2

1$5! = $
A24
B120
C720
D140

Answer: B
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ ${}^6{C_2} = \dfrac{{6 \times 5}}{{2!}} = \dfrac{{6 \times 5}}{{2 \times 1}} = 15$

2${}^6{C_2} = $
A15
B30
C12
D48

Answer: A
Explanation:
${}^6{C_2} = \dfrac{{6 \times 5}}{{2!}} = \dfrac{{6 \times 5}}{{2 \times 1}} = 15$

3If ${}^n{C_3} = 35$, then $n!=?$
A24
B120
C720
D5040

Answer: D

Explanation:
${}^n{C_3} = 35$
$ \Rightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}} = 35$
$ \Rightarrow n\left( {n - 1} \right)\left( {n - 2} \right) = 7 \times 5 \times 3!$
$ \Rightarrow n\left( {n - 1} \right)\left( {n - 2} \right) = 7 \times 6 \times 5$
$ \Rightarrow n = 7$
$\therefore 7! = 5040$

4If $r!=720$ then ${}^8{C_r} = $
A56
B48
C30
D28

Answer: D

Explanation:
$r!=720$ $= 1 \times 2 \times 3 \times 4 \times 5 \times 6$
$ \Rightarrow r = 6$
${}^8{C_6} = {}^8{C_2} = \dfrac{{8 \times 7}}{{2!}} = \dfrac{{8 \times 7}}{{2 \times 1}} = 28$
$\left( {\because {}^n{C_r} = {}^n{C_{n - r}}} \right)$

5${}^8{P_2} = $
A56
B28
C40
D36

Answer: A

Explanation:
${}^8{P_2} = 8 \times 7 = 56$

6${}^7{P_4} = $
A1/3
B1/4
C1/2
D2/3

Answer: C

Explanation:
${}^7{P_4} = 7 \times 6 \times 5 \times 4 = 840$

7${}^n{P_r} = $
A$ \dfrac{{{}^n{C_r}}}{{r!}}$
B$ {}^n{C_r} \times r!$
C$ {}^n{C_r} \times n!$
D$ {}^n{C_r} \times 2n!$

Answer: B

Explanation:
${}^n{P_r} = {}^n{C_r} \times r!$
This is a very important result. Please memorize.

8In how many ways can the letters of the word 'LOVER' be arranged?
A120
B24
C60
D30

Answer: A

Explanation:
All letters of the word are distinct in the word $LOVER$.
$\therefore $ Number of arrangements = $5!$ $= 120$

9In how many ways can the letters of the word 'BEAUTY' be arranged so that the words always start with $A$?
A15
B20
C60
D120

Answer: D

Explanation:
$BEAUTY$ contains $6$ letters.
If we fix the first letter $A$ in the first place,
$\underline A \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} $
Then number of ways of filling Remaining $5$ letters in $5$ places $= 5! = 120$

10In how many ways can the letters of the word 'BEAUTY' be arranged so that the words always start with $A$ and ends with $B$?
A56
B48
C30
D24

Answer: D

Explanation:
$BEAUTY$ contains $6$ letters.
If we fix the first letter $A$ in the first place and $B$ in the last place,
$\underline A \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline B $
Then number of ways of filling Remaining $4$ letters in $4$ places $= 4! = 24$

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Permutations and Combinations: FormulasPermutations and Combinations: Exercise - 1Permutations and Combinations: Exercise - 2