align="absmiddle" alt="\huge\sf\underline{\pink{A}\orange{N}\blue{S}\red{W}\green{E}\purple{R}}" class="latex-formula" ID="TexFormula1" src="https://tex.z-dn.net/?f=%5Chuge%5Csf%5Cunderline%7B%5Cpink%7BA%7D%5Corange%7BN%7D%5Cblue%7BS%7D%5Cred%7BW%7D%5Cgreen%7BE%7D%5Cpurple%7BR%7D%7D" TITLE="\huge\sf\underline{\pink{A}\orange{N}\blue{S}\red{W}\green{E}\purple{R}}">

This question can be answered in several enumerating ways by using principle of counting, However , I would like to suggest the one which uses a simple result of partitioning of a set into a no. of subsets of it . This question may be thought of “ finding the no. of onto (subjective) maps from a set A of n (= 5 here ) elements to the set B of k ( = 3 here ) elements “. Each such maps sends an element a € A to an element x € B, LEAVING no element left un- mapped in B . Further an onto - map from a set A of n elements to a set B of k elements may be characterised by a partition of A into k subsets S(n, k) ( known as Stirling nos. of 2nd kind ) when multiplied with k! gives the required result. That is ;

S( n, k ) .( k!) = C(k, 0)k^n - C(k, 1)(k - 1)^n + C(k, 2)(k -2)^n - …. .. + (-1)^(k-1) .(1^n).

Here in our case n = 5 & k = 3 . Put these values in r.h.s. expression, we get ;

= C(3,0)(3^5) - C(3,1)(2^5) + C(3,2)(1^5) - C(3,3)(0^5) = (3^5) - 3.(2^5) + 3.(1^5)

= 243 - 96 + 3 = 150 , which are the required no. of ways of DISTRIBUTING 5 chocolates to 3 CHILDREN.

#zeaL❤