Question

In how many ways can 6 different rings be put on 5 fingers with at least one ring in each finger.
A. $2400$
B. $3600$
C. $1440$
D. $7200$

Answer 1

Here we solve using the combination formula but keeping in mind that all fingers should have at least one ring on them which means fixing one ring to one finger and deciding for the remaining rings. So, the number of rings we have to allot to fingers gets deducted by one after every step, and since the number of rings is just one more than the number of fingers, therefore we choose for the last remaining ring separately.

• Combination is used when we have to find ways to choose without keeping in mind the order. Then number of all combinations of all $n$ things taken $r$ at a time is given by $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$

Complete step by step answer:
Given, total number of rings is $6$
Number of fingers is $5$
Therefore, we have $5$ places empty to be filled with $6$rings.
For first finger
Number of rings is $6$ and we have to choose one,
Therefore, ways to fill first finger is $^6{C_1}$
For second finger
Now, one ring is occupied.
Number of rings is $5$ and we have to choose one,
Therefore, ways to fill second finger is $^5{C_1}$
For third finger
Now, two rings are occupied.
Number of rings is $4$ and we have to choose one,
Therefore, ways to fill third finger is $^4{C_1}$
For fourth finger
Now, three rings are occupied.
Number of rings is $3$ and we have to choose one,
Therefore, ways to fill fourth finger is $^3{C_1}$
For fifth finger
Now, four rings are occupied.
Number of rings is $2$ and we have to choose one,
Therefore, ways to fill fifth finger is $^2{C_1}$
Total number of ways to fill $5$ fingers with at least one ring on them is given by
$^6{C_1}{ \times ^5}{C_1}{ \times ^4}{C_1}{ \times ^3}{C_1}{ \times ^2}{C_1}$
$= \dfrac{{6!}}{{(6 - 1)!1!}} \times \dfrac{{5!}}{{(5 - 1)!1!}} \times \dfrac{{4!}}{{(4 - 1)!1!}} \times \dfrac{{3!}}{{(3 - 1)!1!}} \times \dfrac{{2!}}{{(2 - 1)!1!}}$
Now from the definition of factorial of a number $p! = p(p - 1)!$
$= \dfrac{{6 \times 5!}}{{5!1!}} \times \dfrac{{5 \times 4!}}{{4!1!}} \times \dfrac{{4 \times 3!}}{{3!1!}} \times \dfrac{{3 \times 2!}}{{2!1!}} \times \dfrac{{2 \times 1!}}{{1!1!}}$
Cancelling out the same terms from numerator and denominator and substituting $1! = 1$
$= 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$= 720$
Now, all five fingers have at least one ring on them. So, we are left with one ring that can be put on any of the five fingers. Therefore, the number of ways to put on the last ring is $5$.
Therefore, the total number of ways in which $6$ different rings are put on $5$ fingers with at least one ring in each finger $= 720 \times 5 = 3600$.

Therefore, option B is correct.

Note:
Students are likely to make mistakes when moving on to the next allotment of the ring, usually students don’t deduct the number of rings which gives wrong calculations. Generally, students make mistake of substituting values of $n = 6$ and $r = 5$ which gives $^n{C_r}{ = ^6}{C_5} = \dfrac{{6!}}{{5!1!}} = 6$ which is wrong.