SOLUTION: "In how many ways can 7 students be divided into 3 unordered groups such that one group contains 1 student and the other two groups contain 3 students each?"

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Question 1209238: "In how many ways can 7 students be divided into 3 unordered groups such that one group contains 1 student and the other two groups contain 3 students each?"

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52792) About Me  (Show Source):
You can put this solution on YOUR website!
.

First group of one student can be selected from 7 students in C%5B7%5D%5E1 = 7 ways.

Second group of 3 students can be selected from remaining 7-1 = 6 students in C%7B%5B6%5D%5E3 = %286%2A5%2A4%29%2F%281%2A2%2A3%29 = 5*4 = 20 ways.


Third group of 3 students will consist of 3 remaining students, so there is only one choice to select this group.



So, the ANSWER  to the problem's question is this product

    in how many ways = 7*20 = 140 different ways.

Solved.



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


It is instructional to the student to see that the three groups can be formed in any order.

(1) Choose the "group" of 1 student first

Choose the 1 student from among the 7 in C(7,1) = 7 ways
Choose either group of 3 students from among the remaining 6 students in C(6,3) = 20 ways
Choose the last group of 3 students from among the remaining 3 students in C(3,3) = 1 way

Total number of ways: 7*20*1 = 140

(2) Choose the "group" of 1 student second

Choose 3 students from among the 7 in C(7,3) = 35 ways
Choose the "group" of 1 student from among the remaining 4 students in C(4,1) = 4 ways
Choose the last group of 3 students from among the remaining 3 students in C(3,3) = 1 way

Total number of ways: 35*4*1 = 140

(3) Choose the "group" of 1 student last

Choose 3 students from among the 7 in C(7,3) = 35 ways
Choose the second group of 3 students from among the remaining 4 students in C(4,3) = 4 ways
Choose the last "group" of 1 student from among the remaining 1 student in C(1,1) = 1 way

Total number of ways: 35*4*1 = 140