Question

Number of linear arrangements of $4$ alike of one kind and $5$ alike of another kind is NOT equal to

• A. number of $5-$digit numbers whose digits are in increasing order from left to right.
• B. number of linear arrangements of $8$ identical apples and $4$ identical oranges if no two oranges are together.
• C. number of ways of distribution of $9$ distinct toys between two children when one is having exactly one toy more than the other.
• D. number of ways in which $8$ students can be divided into two teams, that need not necessarily be of equal size.

Answer 1

Answer: (C), (D)

number of ways in which $8$ students can be divided into two teams, that need not necessarily be of equal size.

Number of linear arrangements of $4$ alike of one kind and $5$ alike of another kind $=\frac{9!}{4!5!}={}^9{C}_4$

For increasing order : We have only $9$ options to choose from, as including $0$ and arranging it in increasing order will put $0$ on the front and with $0$ on front, the number will be a $4-$digit number.
So, for this case, we have ${}^9{C}_4$ ways.

Using gap method : We have $9$ gaps between $8$ apples and we have to place $4$ oranges between them. It is equal to number of ways to select any $4$ out of $9$ distinct objects, that is equal to ${}^9{C}_4$

One child gets $4$ and another gets $5$ toys.
We need to distribute $4$ toys to any one. Then another child will get remaining $5$ automatically.
Number of ways $={}^9{C}_4$
As both the children are different, so they can exchange their toys.
Hence, required number of ways $=2\times {}^9{C}_4$

We have to select only one team because another team will get selected automatically from the remaining students.
Required number of ways $={}^8{C}_1+{}^8{C}_2+{}^8{C}_3+\frac{{}^8{C}_4}{2}=8+28+56+35=127\ne {}^9{C}_4$
$(\because {}^9{C}_4=126)$