Basic Concept of Permutations and Combinations-Test 2

given word ALLAHABAD

no.of ways =n!/p!q!

total no.of letter n=9

no of A   p=4

no of L   q=2

no of words =9!/4!x2!=9x8x7x6x5x4!/4!x2!=7560

No.of parallelogram are formed from a set of m parallel lines intersecting another set of n

$$\begin{gathered} If 10C_3+2.10C_4+10C_5=n{C}_5 \\ (10C_3+10C_4)+(10C_4+10C_5)=n{C}_5 \\ 10+1C_5+10+1C_4=n{C}_5 \\ 11C_5+11C_4=n{C}_5 \\ 11+1C_5=n{C}_5 \\ 12C_5=n{C}_5 \\ n=12 \end{gathered}$$

1       TH       H          T         U

2       6x       5          x4        x3     =360

3

4

5

6

no.of numbers between 1000 and 10000 using the digit 1,2,3,4,5,6=360

$$\begin{gathered} The maximum no.of points of intersection of n circle \\ =n{p}_2 \\ =10p_2 \\ =10x9 \\ =90 \end{gathered}$$

Since 2 particular books are to be kept always at the first and last place ,so if we fix places,the remaining 5 books can be arranged in 5! ways

those,2 books can also change their places in 2! ways

the total number of arrangements are=5!x2!=120×2=240 ways

In the first place only M can come.therefore,no.of ways of filling first place is 1!=1 way

In the last place ,any letter other than M and N can come. therefore ,no of ways in which last place can be filled is 4 ways i.e 6-2=4 ways

The remaining 4 places have no restrictions.They can be filled by any of the remaining 4 letters.so,the no.of ways of filling those 4 place are 4! ways

so,the total no.of words that can be formed are=1x4x4! ways =4×24=96 ways

$$\begin{gathered} Number of waysof arranging 17P_{17 } i.e. 17! ways \\ but 7 of them are black ,6 red and 4 white \\ no.of ways of arrenging the balls =\frac{17!}{7!x6!x4!}=4084080 \end{gathered}$$

$$\begin{gathered} no.of ways of selecting 2 gents=4C_2 \\ no.of ways of selecting ladies so that atleast the number of ladies is \\ double of gents=6C4+6C_5+6C_6 \\ total ways =4C_{2}x(6C_4+6C_5+6C_6) \\ =6x(15+6+1) \\ =132 ways \end{gathered}$$

no.of ways of drawing 3 balls at a time=120 ways

no.of ways of drawing 3 white balls out of 5 white balls =10 ways

total no.of ways=favourable cases/total no.of cases=10/129=1/12

$$\begin{gathered} no.of quadrilaterals that can be formed from 6 points on a circle \\ =6C_4 \\ =\frac{6!}{2!4!} \\ =30/2 \\ =15 \end{gathered}$$

let us assume 4 girls to be seated together so they can be arranged in 4! ways

now,if we assume 4 girls to be one single group

total number of ways of arranging boys and girls is 7!

required numbers of ways are 7!x4!

total no.of 2 digits that can be formed=9×8=72

total no.of 3 digits that can be formed=9x8x7=504

total no.of 1 digits that can be formed=9

total numbers that can be formed=9+72+504=585

5 places between the trees

0                   0                0               0                 0

X                     X               X                X               X                   X

T        5           T        4      T       3       T        2      T         1         T

no.of ways 5! =120

$$\begin{gathered} if one particular player is not selected \\ no of ways=14C_{11} \\ =14C_3 \\ =\frac{14x13x12}{3x2x1} \\ =364 \end{gathered}$$

$$\begin{gathered} given , 5 girls + 3 boys \\ required exactly 3 girls \\ probability of 3 girls=\frac{5C_3}{8C_3}=10/56=5/28 \end{gathered}$$

$no.of ways =n^r=2^{12}=4096$

words DRAUGHT

if both vowels may not be separated this mean that both vowels come together

D R G H T     A U

6  5 4  3 2      1

no.of ways if two vowels come together=6!x2!=720×2=1440

$$\begin{gathered} if 13C_6+2.13C_5+13C_4=15C_x \\ (13C_6+13C_5)+(13C_5+13C_4)=15C_x \\ 14C_6+14C_5=15C_x \\ X=6 \\ 15C_6=15C_x \end{gathered}$$

$$\begin{gathered} for shaking hands \\ no of ways =n{C}_2 \\ =10C_2 \\ =10x9/2x1 \\ =45 \end{gathered}$$

$$\begin{gathered} given ,15C_{3r}=15C_{r+3} \\ then ,3r+r+3=15 \\ 4r=12 \\ r=3 \end{gathered}$$

no of different word can be formed from the letter of LIBERTY=7!=5040

no of ways can a family consist of 3 children here different birthday in a leap yr= 366x365x364

no of ways = 3! 3! 5! 3!

=6x6x120x6=25920