How many 4-letter words with or without meaning, can be formed out of the letters of the word

Solution

The number of objects(n), in this case, is 5, as the word SMOKE has 5 alphabets.and r=3, as 3-letter word has to be chosen.Thus, the permutation will be:Permutation (when repetition is allowed) =5×5×5=125

How many 4 - letter words (with or without meaning) containing two vowels can be constructed using only the letters (without repetition) of the word 'LUCKNOW'?

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NDA 02/2021: Maths Previous Year paper (Held On 14 Nov 2021)

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1. 240
2. 200
3. 150
4. 120

Answer (Detailed Solution Below)

Option 1 : 240

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Electric charges and coulomb's law (Basic)

10 Questions 10 Marks 10 Mins

Concept:

If n is a positive integer and r is a whole number, such that r < n, then P(n, r) represents the number of all possible arrangements or permutations of n distinct objects taken r at a time.

It can be represented as nPr = \(\frac{n!}{(n-r)!}\).

The combination is defined as “An arrangement of objects where the order in which the objects are selected does not matter.” 

nCr = \(\frac{n!}{r!(n-r)!}\) ,  when n < r

Where n = distinct object to choose from

C = Combination

r = spaces to fill

Calculation:

Vowels = 2

Consonants = 5

Total Alphabets = 7

Since 4 letter words must include 2 vowels, we don't need to select them, and the rest of the 2 letters will be taken from 5 consonants.

Number of ways of selecting 2 letters from 5 consonants = 5C2 = 10

Arrangement of all 4 letters will be given by 4! = 24 ways

Total number of arrangements = 5C2 × 4! = 10 × 24 = 240 ways

∴ The total number of words that can be formed is 240.

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11.	In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
Answer: Option A Explanation: Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = 7 x 6 x 3 = 63. 2 x 1

13.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A. 10080 B. 4989600 C. 120960 D. None of these Answer: Option C Explanation: In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. Number of ways of arranging these letters = 8! = 10080. (2!)(2!) Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4! = 12. 2! Required number of words = (10080 x 12) = 120960.

How many 4

How many 4