# Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. What is the total number of ways in which Dishu can distribute $9$ distinct gifts among his $8$ distinct girlfriends such that each of them gets at least one gift?
 ✔ A. $72 × 8!$ ✖ B. $144 × 8!$ ✖ C. $36 × 8!$ ✖ D. $9!$

Solution:
Option(A) is correct

As every girl friend should get one gift.

The number of ways $8$ distinct gifts can be selected is: ${^9C_8} = 9$ ways.

The number of ways each GF gets one gift each out of these $8$ selected gifts $8!$

Total number of ways $8$ gifts can be distributed is $9 × 8!$.

Now the last one gift can be given to any of the $8$ GF hence the total number of ways of

distributing,

$= 9 × 8! × 8$

$= \textbf{72 × 8! ways.}$

Edit: Based on the comments final answer has been changed to $\textbf{72 × 8! ways}$ from $\textbf{36 × 8! ways.}$

## (8) Comment(s)

Max
()

Fairly certain the answer is instead ${^9P_8} + {^9C_2}\times 8!$.

If each girlfriend gets one distinct gift that is a permutation 9 factorial. add that with the chance of him getting 2 gifts for one girl and one for the rest.

This is done ${^9C_2} \times 7!$ ways for one girl, and for all of them would be ${^9C_2} \times 7! \times 8$

Suyog
()

why to divide it by 2?

Abhishek Anand
()

Why to divide it by 2?

Siddharth Singh
()

one among 8 gfs will get 2 gifts and remaining 7 will get one each.

so total of 9 gifts will be distributed among 8 gfs.

i.e; 11111112.

Gf who will get 2 gifts can be find out in 8c1 ways = 8 ways.

now now 2 gifts can be given to selected gf in $^9c_2$ ways. and remaining 7 gifts can be given to remaining 7 gf in 7! ways.

so total no of ways $= 8*^9c_2*7!$

$=\dfrac{8*(9*8)}{2}*7!$=36*8*7!=36*8!$Rahul () Why it divided by 2... It can be${^9C_8} * 8! * 8$Suribabu () u mad bro?? Ravi () I think the answer must be$72*8!\$..

Happy
()

why divide by 2 in this question