\[\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}\]
\[=\frac{\left(4\times 5\right)+\left(5\right)+\left(1\right)}{5!}\]
\[=\frac{20+5+6}{5\times 4\times 3\times 2\times 1}\]
\[=\frac{26}{120}\]
\[=\frac{13}{60}\]
In how many ways can 5 apples be distributed among 4 boys, there being no restriction to the number of apples each boy may get?
The number of permutations of n different things taken r at a time when each thing may be repeated any number of times is \[[n^r]\].
Here
Hence, the required number of ways
\[=n^r=4^5\]
Find the number of ways in which 5 different beads can be arranged to form a necklace.
Number of circular arrangements of n different things when clockwise and anticlockwise arrangements are not different
\[=\frac{\left(n-1\right)!}{2}\]
Here n = 5
Hence, required number of arrangements
\[=\frac{\left(n-1\right)!}{2}\]
\[=\frac{\left(5-1\right)!}{2}\]
\[=\frac{4!}{2}\]
\[=\frac{4.3.2}{2}\]
\[=4.3\]
\[=12\]
In a class, there are 20 boys and 15 girls. The teacher wants to select 1 boy and 1 girl to represent the class for a function. In how many ways can the teacher make this selection?
Here the teacher is to perform two operations:
Now, By the fundamental principle of counting, the required number of ways is
\[20 × 15 = 300\].
In how many ways a committee consisting of 3 men and 2 women, can be chosen from 7 men and 5 women?
Hence , the committee can be chosen in \[7_{C_3} × 5_{C_2} = 350\] ways.
In how many ways can 5 members forming a committee out of 10 members be selected so that two particular members must be included.
According to the question, we have to select
(5 - 2) = 3 members out of (10 - 2) = 8 members
\[\therefore\] reuired number of ways \[= 8_{C_3} = 56\] ways.
Formula - Number of the combination of n different things taken r at a time when P particular things are always included
\[= n-p_{C_{r-p}}\]
Here
Hence, required number of ways
\[= n-p_{C_{r-p}}\]
\[= 10-2_{C_{5-2}}\]
\[= 8_{C_{3}}\]
\[= \frac{8!}{3!5!}\]
\[= \frac{8\times7\times6}{6}\]
\[=56\] ways.
In how many ways can 5 members forming a committee out of 10 members be selected so that two particular members must not be included.
According to the question, we have to select
5 members out of (10 - 2) = 8 members
\[\therefore\] reuired number of ways \[= 8_{C_5} = 56\] ways.
Formula - Number of the combination of n different things taken r at a time when P particular things are never included
\[= n-p_{C_r}\]
Here
Hence, required number of ways
\[= n-p_{C_r}\]
\[= 10-2_{C_5}\]
\[= 8_{C_{3}}\]
\[= \frac{8!}{5!3!}\]
\[= \frac{8\times7\times6}{6}\]
\[=56\] ways.
In a party every person shakes hands with every person. If there was a total of 105 handshakes in the party, find the number of persons who were present in the party.
In a party, every person shakes with every other person. If there was a total of H handshakes in the party, then the number of persons n who were present in the party can be calculated from the equation
\[\frac{n\left(n-1\right)}{2}=H\]
Here H = 105
\[\therefore \frac{n\left(n-1\right)}{2}=105\]
\[\implies n(n-1)=15 \times (15-1)\]
\[\implies n=15\].
How many straight lines can be drawn with 18 points on a plane of which no points are collinear?
There are n points in a plane and no points are collinear, then the number of straight lines that can be drawn using these n points are given by
\[\frac{n(n-1)}{2}\]
Here n = 18
\[\therefore\] required number of lines
\[=\frac{18(18-1)}{2}\]
\[=\frac{18\times17}{2}\]
\[=\frac{306}{2}\]
\[=153\]
There are 14 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points as vertices.
If there are n points in a plane out of which m points are collinear. The number of triangles formed by the points as vertices are given by :
\[n_{C_3-}m_{C_3}\]
Here
\[\therefore\] required number of triangles
\[=14_{C_3}-4_{C_3}\]
\[=364-4\]
\[=360\]
In a chess board, there are 9 vertical and 9 horizontal lines. Find the number of rectangles formed in the chess board.
If there are m horizontal lines and n vertical lines then the number of different rectangles formed are given by
\[m_{C_2}\times n_{C_2}\]
Here
\[\therefore\] required number of rectangles
\[=9_{C_2}\times 9_{C_2}\]
\[=36\times 36\]
\[=1296\]
Find the number of triangles formed by joining the vertices of an octagon.
The number of triangles which can be formed joining the angular points of a polygon of n sides as vertices are
\[\frac{n(n-1)(n-1)}{6}\]
Here n = 8
\[\therefore\] required number of triangles
\[=\frac{8(8-1)(8-1)}{6}\]
\[=\frac{8\times7\times6}{6}\]
\[=56\]
How many diagonals are there in a decagon?
The number of diagonals which can be formed by joining the vertices of a polygon of n sides are
\[\frac{n(n-3)}{2}\]
Here n = 10
\[\therefore\] required number of diagonals
\[=\frac{10(10-3)}{2}\]
\[=\frac{10\times7}{2}\]
\[=35\]
Find the number of triangles that can be formed with 14 points in a plane of which no three points are colinear.
If there are n points in a plane and no three points are collinear, then the number of triangles formed with n points are given by
\[\frac{n(n-1)(n-2)}{6}\]
Here n = 14
\[\therefore\] required number of triangles
\[\frac{14(13-1)(12-2)}{6}\]
\[=364\]
Find the number of quadrilaterals that can be formed by joining the vertices of a septagon.
The number of quadrilaterals that can be formed by joining the vertices of a polygon of n sides are given by
\[\frac{n(n-1)(n-2)(n-3)}{24}\]
Here n = 7
\[\therefore\] required number of quadrilaterals
\[=\frac{n(n-1)(n-2)(n-3)}{24}\]
\[=\frac{7(7-1)(7-2)(7-3)}{24}\]
\[=\frac{7\times6\times5\times4}{24}\]
\[=35\]