Name: Number of Onto Functions
Uploaded: 2019-07-19
Duration: 9 min 3 s
Description: In this video, I'll show you how to prove the number of onto functions formula. This will prove handy for a better understanding of this concept.

Question

There are counters available in

7

different colours. Counters are all alike except colour and they are atleast ten of each colour. Find the number of ways in which an arrangement of

10

counters can be made. How many of these will have counters of each colour?

Accepted Answer

For each of the $10$ counters, there are $7$ choices of colours to be selected from. So by principle of multiplication, total such arrangements will be $7^{10}$ .

For having a counter of each colour, three cases are possible:

Case I: $4$ of one colour and remaining $6$ of remaining $6$ colours.

In ${}^{7}C_{1}$ ways, the colour can be selected for which we have $4$ counters (which will be identical also), the remaining can be selected in ${}^{6}C_{6}$ ways and correspondingly total arrangements $= {}^{7}C_{1} \times {}^{6}C_{6} \times \frac{10!}{4!}$ .

Case II: $3$ of one colour, $2$ of another colour and remaining $5$ of remaining $5$ colours.

In ${}^{7}C_{2} \times {}^{2}C_{1}$ ways, the first two conditions will be met and the remaining can be selected in ${}^{5}C_{5}$ ways. Correspondingly total arrangements $= {}^{7}C_{2} \times {}^{2}C_{1} \times {}^{5}C_{5} \times \frac{10!}{3! 2!}$

Case III: $2$ of one colour, $2$ of second colour, $2$ of third colour and remaining $4$ of remaining $4$ colours.

This selection can be made in ${}^{7}C_{3} \times {}^{4}C_{4}$ ways and correspondingly total arrangements $= {}^{7}C_{3} \times {}^{4}C_{4} \times \frac{10!}{2! 2! 2!}$

So, total arrangements having counter of each colour $= {}^{7}C_{1} \times {}^{6}C_{6} \times \frac{10!}{4!} + {}^{7}C_{2} \times {}^{2}C_{1} \times {}^{5}C_{5} \times \frac{10!}{3! 2!} + {}^{7}C_{3} \times {}^{4}C_{4} \times \frac{10!}{2! 2! 2!} = \frac{49}{6} \times 10!$

There are counters available in $7$ different colours. Counters are all alike except colour and they are atleast ten of each colour. Find the number of ways in which an arrangement of $10$ counters can be made. How many of these will have counters of each colour?

Important Questions on Permutation and Combination

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