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In how many ways can 7 students be assigned to 1 triple and 2 double hotel rooms during a conference?

  • Last Updated : 16 Dec, 2021

Given a definite number of elements, their different arrangements, either by choosing one after the other, or some of them, or all of them at a time are called permutations. They are regarded as the process of assigning a linear sequence to the members of a given sequence. It is also referred to as the process of re-ordering the elements of a given sequence or series.

In other words, to permutate a sequence means to list out all possible arrangements of that sequence.

For example, the series {1, 2} can be written in two ways: either as {1, 2} or {2, 1}.

Number of permutations when r number of elements are arranged out of n elements in a given sequence

nPr = n!/(n-r)!

For example, let n = 5 {A, B, C, D, E} and r = 2 (all permutations having two elements). The answer is 5P2 = 5! /(5-2)! = 20. The 20 permutations are: AB, AC, AD, AE, BA, BC, BD, BE, CA, CB, CD, CE, DA, DB, DC, DE, EA, EB, EC and ED.

Combinations

It is defined as the process of selecting either one or two or some elements from a given sequence, regardless of the order of the elements. Say, two elements of a series are to be chosen which only has 2 elements, to begin with, then the order of those elements won’t matter.

Number of combinations when r number of elements are selected out of n elements in a given sequence,

nCr = n!/r!(n-r)!

For example, let n = 5 and r = 2, then number of ways to select 2 elements out of 5 = 5C2 = 5!/2! (5-2)! = 10.

In how many ways can 7 students be assigned to 1 triple and 2 double hotel rooms during a conference?

Solution:

This problem can be interpreted as having to put the 7 students into groups of 3, 2 and 2.

Number of ways to choose 3 students in the triple = 7C3 = 7! / 3!4! = 35  

Number of ways to choose 2 out of the remaining 4 students = 4C2 = 4!/ 2!2! = 6

Number of ways to choose 2 students out of the remaining two students = 1

Total number of arrangements = 35 × 6 × 1 = 210.

Hence, 7 students can be assigned to 1 triple and 2 double hotel rooms during a conference in 210 ways.

Similar Problems

Question 1. In how many ways can five persons are to be selected to form a committee so that at least 3 men are there on the committee from a group of 7 men and 6 women?

Solution:

At least three men on the committee means we can have either exactly three, four or all five men in the committee.

Required number of ways = (7C3 × 6C2) + (7C4 × 6C1) + (7C5)

\frac{7×6×5}{3×2×1}  + \frac{7×6×5}{3×2×1} × 6   + \frac{7×6}{2×1}

= 525 + 210 + 21

= 756

Question 2. In how many ways can the letters of the word ‘LEADING’ be arranged so that the vowels always come together?

Solution:

There are 3 vowels in the word. Number of ways to arrange these vowels among themselves = 3! = 6

Now as for the 4 letters, number of ways to arrange = 5! = 120

Total number of ways of arranging the letters = 120 × 6 = 720.

Question 3. Out of 8 consonants and 5 vowels, how many words of 4 consonants and 3 vowels can be formed?

Solution:

Number of ways of selecting 4 consonants out of 8 and 3 vowels out of 5 = 8C4 × 5C3

\frac{8 ×7 ×6 ×5 ×4!}{4!  × 4!}  × \frac{5 ×4 ×3!}{3!  × 2!}

= 70 × 10 = 700

Number of ways of arranging the 7 letters among themselves = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Number of words that can be formed = 5040 × 700 = 3528000.

Question 4. How many four-letter words can be made out of the word ‘LOGARITHMS’ if repetition is not allowed?

Solution:

Since there are 10 different letters in the word ‘logarithms’.

Required number of words = 10P4

= 10!/6! 

= 5040

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