Question

Number of shortest way in which we can reach from the point $(0,0,0)$ to point $(3,7,11)$ in space where the movement is possible only along the x-axis, y-axis and z-axis or parallel to them and change of axes is permitted only at integral points (an integral point is one which has its coordinate as integer) is

• A. Equivalent to number of ways of dividing $21$ different objects in three groups of size $3,7,11$
• B. Equivalent to coefficient of $y^3{z}^7$ in the expansion of $(1+y+z{)}^{21}$
• C. Equivalent to number of ways of distributing $21$ different objects in three boxes of size $3,7,11$
• D. Equivalent to number of ways of arranging $21$ objects of which $3$ are alike of one kind, $7$ are alike of second type, and $11$ are alike of third type

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Equivalent to number of ways of dividing $21$ different objects in three groups of size $3,7,11$
B Equivalent to coefficient of $y^3{z}^7$ in the expansion of $(1+y+z{)}^{21}$
C Equivalent to number of ways of distributing $21$ different objects in three boxes of size $3,7,11$
D Equivalent to number of ways of arranging $21$ objects of which $3$ are alike of one kind, $7$ are alike of second type, and $11$ are alike of third type

There are 3 units of distance to be covered along X, 7 along Y and 11 along Z.
It can be thought of a word: $XXXYYYYYYYZZZZZZZZZZZ$
Number of distinct words that can be formed is the same as the different routes that can be formed to reach from (0, 0, 0) to (3, 7, 11) is: $\frac{21!}{3!7!11!}$
Option (a): 21 different objects may be divided in 3 groups having 3, 7 and 11 objects in: ${}^{21}{C}_3{\times }^{18}{C}_7{\times }^{11}{C}_{11}=\frac{21!}{3!7!11!}$
Option (b): coefficient of $y^3{z}^7$ in ${(1+y+z)}^{21}$ i.e. coefficient of $1^{11}{y}^3{z}^7$ in ${(1+y+z)}^{21}$ $=\frac{21!}{3!7!11!}$
Option (c): 21 different objects may be divided in 3 boxes having 3, 7 and 11 objects without considering the order in which they are filled up ${=}^{21}{C}_3{\times }^{18}{C}_7{\times }^{11}{C}_{11}=\frac{21!}{3!7!11!}$
Option (d): number of ways of arranging 21 objects of which 3 are alike of one kind, 7 are alike of second type, and 11 are alike of third type = $\frac{21!}{3!7!11!}$
Hence, (a), (b), (c), (d) are correct.