Question

There are 30 pupils in a class. (a) You make a row of 7 pupils. In how many ways can this be done? (b) You divide the class into two groups of 15 each. In how many ways can this be done? (c) You give each of the 30 pupils one type of cooldrink from 5 different types of cooldrink. In how many ways can this be done? (d) Ignoring who gets which cooldrink, how many different cooldrink combinations are possible if you choose 30 cooldrinks from 5 types?

Answer 1

(a) Out of 30 pupils 7 pupils can be selected in $^{30}C_7$ ways

  and these 7 selected can be arranged in 7! ways-

 

 Total number of ways$=^{30}C_7\times 7!=30*29*28*27*26*25*24$

(b)Number of ways in which we can divide $= ^{30}C_{15}$

(c) Let the five type of cooldrinks be A,B,C,D and E.

$1^{st}$ pupil has a choice to select any one of the cooldrink out of these 5 cooldrinks in $^{5}C_1$ ways =5

Similarly all the remaining 29 pupils have the same ways i.e. 5

So Total number of ways that It can be done$= 5\times 5^{29}=5^{30}$

(d) Let the cooldrinks types be A,B,C,D and E.

  Let us take that $x_1,x_2,x_3,x_4,x_5$ number of cooldrinks are selected of type A,B,C,D and E respectively.

 According to case of multinomial theorem-

     $0\le x_2\le 30\\0\le x_2\le 30\\0\le x_3\le 30\\0\le x_4\le 30\\0\le x_5\le 30$

and $x_1+x_2+x_3+x_4+x_5=30~~~~~~~-(1)$

Number of non-negative integral solution to equation (i) $= ^{n+r-1}C_r$

      SO here, n=5 and r=30

So, $= ^{5+30-1}C_30=^{34}C_{ 30}=46376 ways$