3. How many 5 letter wrd can be frmed using MANAGEMENT such that if any 2 alike letters are there, then they are always together.
Answers
Answer:
In the word “management” there are
2 “m”s ; 2 “a”s; 2 “n”s ; 2 “e”s ; 1 “g”, and 1 “t”
So creating a 5 letter word can have any of those letters in combination, the additional constraint being that whenever any letter repeats, they occur together.
We can divide the problem into following cases:
1. No repetition
We choose 0 from the 4 letters to repeat. Then from the remaining 6 options we choose 5, and permute them.
This is simply (⁴C₀)∗(⁶C₅)∗5!
as we have 6 letters to choose from and 5 letter words.
2. One repetition allowed
We can choose one from 4 letters which can be repeated. Now we need to choose 3 more letters as the previously chosen letter will be repeated twice. So we have 5 options. After choosing we permute them while keeping the repeated ones together, so we only have 4 “items” to permute. The count is
This is simply (⁴C₁) × (⁵C₃) × 4!
3. Two repetitions allowed
Simply choose two letters to repeat, and choose one letter from the rest, and permute them. Since you have to keep the repeated ones together, you have only three “items” to permute. The total count is
This is simply (⁴C₂) × (⁴C₁) × 3!
∴ The total count of words is
= (⁴C₀) × (⁶C₅) × 5! + (⁴C₁) × (⁵C₃) × 4! + (⁴C₂) × (⁴C₁) × 3!
= 720 + 960 + 144
= 1824
Step-by-step explanation: