Question

in how many ways 15students be divided into
three groups
​

Answer 1

Step-by-step explanation:

15 students and 3 groups of equal size; i.e., 5 students per group.

Now, any 2 students from 15 can be chosen in (15C2) = 105 ways.

So, this group (say, group 1) still needs 3 more from remaining 13; this is possible in (13C3) = 286 ways.

Next, group 2 of 5 students can be formed from remaining 10 in (10C5) = 252 ways.

Thereafter, group 3 of 5 students can be formed from remaining 5 in (5C5) = 1 way.

Now, as per my interpretation, groups 1, 2 & 3 should be non-distinguishable; so permutation among them is unnecessary.

So, the number of ways 15 students are to be divided into 3 equal groups, so that 2 particular students are always together = 105*286*252*1 = 7567560 ways.

However, if anybody interprets groups 1, 2 & 3 are distinguishable through different dresses or by naming them as separate houses (Tagore house, Raman house & Viveknanda house); these 3 groups become open to permutation; and in this case the answer will be (3!)*7567560 = 45405360 ways.

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Answer 2

Given: 15 students must be divided into three groups.

To find: Number of ways in which that can be done.

Solution:

• Combinations denote the number of ways of selecting a number of elements taken at a time.
• It can be represented and calculated as,

        $^{n} C_{r} = \frac{n!}{(n-r)! r!}$

• Here, n is the number of elements and r is the number of elements taken at a time.

• In the given question, 15 is the number of elements and 3 is the number of elements taken at a time.

        $^{15} C _{3} = \frac{15!}{12! 3!}$

                 $= 455$

Therefore, the number of ways in which 15 students can be divided into three groups is 455.