Basic Concept of Permutations and Combinations-Test 1

$$\begin{gathered} The number of triangles that can be formed from a set of 12 points \\ =12C_3 \\ \sin ce 7 points are on the same line,therefore no triangle can be formed from these points \\ i.e.number of triangles=12C_3–7C_3=220–35=185 \end{gathered}$$

The number of ways of filling first two places with english alphabets=26×25=650

The number of ways of filling last two places with distinct numbers=9×8=72

The number of code words that can be formed are=650×72=46800

There are two cases possible:

CASE 1:- When mathematics part-II is borrowed(i.e it means part-I has also been borrowed )

CASE 2:-when mathematics part-II is not borrowed(i.e.3 3 books are to be selected out of 7)

hence,total number of ways =35+6=41 ways

The candidate can select 8Q by selecting at least 3 from each part in the following ways :

hence,the total number of ways in which the candidate can select the question will be=35+175+210=420 ways

$$\begin{gathered} majority decision can be taken by a bench of 5 judges in \\ =5C_3+5C_4+5C_5=10+5+1=16 ways \end{gathered}$$

$$\begin{gathered} p(7,k)=60p(7,k–3) \\ 7p_k=60x7p_{k–3} \\ \frac{7!}{(7–k)!}=60x\frac{7!}{[7–(k–3\left)\right]!} \\ \frac{1}{(7–k)!}=60x\frac{1}{[7–(k–3)!} \\ \frac{(10–k)!}{(7–k)!}=60 \\ \frac{(10–k)(9–k)(8–k)(7–k)!}{(7–k)!}=60 \\ {k}^3–27k^2+24k–660=0 \\ (k–5)(k^2–22k+132)=0 \\ k=5,\sin ce k^2–22k+132=0 gives imaginary roots \end{gathered}$$

The word FAILURE have 7 letters out of which 3 letters F,L and R are consonants and 4 letters A,I,U and E are vowels.

The four positions to be filled up with consonants are indicated below:

1      2       3       4       5         6        7

F               L                R              not filled

Since,there are only 3 consonants ,the total number of permutations are

Again,for the arrangement described above ,the four vowels can occupy the 4 remaining positions not occupied by consonants =4!=24 ways

hence the total number of arrangements are =24×24=576

$$\begin{gathered} Total number of trials =5C_2=10 ways \\ no.of trials for no light in the room =3C_2=3 \\ The room shall be lighted in=10–3=7 ways \end{gathered}$$

The no.of ways in which 4 men can be seated at the circular table so that there is a vacant seat between every pair of men is=(4-1)!=3!=6 ways

hence,no.of ways in which 4 vacant seats can be occupied by 4 women =4!=24 ways

Required no of ways =6×24=144 ways

$$\begin{gathered} \sum _{r=1}^{5}5C_r=5C_1+5C_2+5C_3+5C_4+5C_5 \\ =5+10+10+5+1=31 \end{gathered}$$

$$\begin{gathered} 6P_r=24x6C_r \\ \frac{6!}{(6–r)!}=24x\frac{6!}{r!x(6–r)!} \\ 4!=\frac{24}{r!} \\ r!=\frac{24}{4!} \\ r!=4! \\ r=4 \end{gathered}$$

$There are 4 odd places and 2 letters,hence this can be done in =4P_2=12 ways$

The remaining 6 letters i.e.O,R,I,N.T,L can be arranged in 6! ways

hence,the total no.of arrangements=6!x12=8640

$$\begin{gathered} 1000C_{98}=999C_{97}+X{C}_{901} \\ \sin ce ,n+1C_r=n{C}_r+n{C}_{r–1} \\ 999C_{98}=999C_{97}+X{C}_{901} \\ 999C_{98}+999C_{97}=999C_{97}+X{C}_{901} \\ 999C_{98}+999C_{97}=999C_{97}+X{C}_{98} \\ X=999 \end{gathered}$$

total no.of numbers that can be formed from the digits 4,5,5,0,4,5,3=7!/2!x3!=420 ways

out of these 420 numbers some will begin with 0 and are less than one million ,so they are to be rejected

number of numbers beginning with

0=6!/2!x3!=60 ways

hence required number of numbers=420-60=360 ways

since there is no regard for order,the contractor can select any of the 3 helpers out of 10 men.

this can be done in 

$$\begin{gathered} No.of way\sin which these balls can be arranged \\ =\frac{12!}{3!x4!x5!} \\ =27720 ways \end{gathered}$$

The no.of ways in which 6 articles clerks can be selected out of 10 candidates

Using the given restrictions ,we must have AB or AC and BC or BD.

therefore,we have the following alternatives:

(i)ABC,D,E,F which gives (4-1)! or 31 ways

(ii)ABD,C,e,f which gives (4-1)! or 31 ways

(iii)AC,BD,E,F which gives (4-1)! or 31 ways

hence,the total number of ways are =3!+3!+3!=6+6+6=18 ways

$$\begin{gathered} n{p}_r=n{p}_{r+1}\Rightarrow n=r+1...........\left(1\right) \\ n{c}_r=n{c}_{r–1}\Rightarrow n=2r–1..........\left(2\right) \\ from 1\&2 we get \\ r+1=2r–1 \\ r=2 \\ hence,n=2+1=3 \end{gathered}$$

number of ways of painting a face of a cube by 6 colours is 6,since any of the 6 colours can be used to paint the face of the cube.