can anyone solves it? please . webwork / 18 fa 381p / hw5-counting /4 HW5 - Coun

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can anyone solves it? please .

webwork / 18 fa 381p / hw5-counting /4 HW5 - Counting: Problem 4 Previous Problem ListNext (1 point) 4 -letter words" are formed using the letters A, B, C, D, E. F. G. How many such words are possible for each of the folowing conditions? (a) No condition is imposed Your answer is: (b) No letter can be repeated in a word Your answer is : (c) Each word must begin with the letter A Your answer is (d) The letter C must be at the end. Your answer is : (e) The second letter must be a vowel. Your answer is: Note: You can earn partial credit on this problem. Submit Answers Preview My Answers

Explanation / Answer

Solution:

There are 7 letters which are A, B, C, D, E, F, G

Solution(a)

No condition is imposed than no. Of ways that the four letter word made = 7C1 × 7C1 × 7C1 ×7C1 = 7×7×7×7 = 2401 words

So there is 2401 words can be made for four letter of there is no condition imposed.

Solution(b)

No letter can be repeated in a word than no. Of ways word can be made for four letters

=7C1 ×6C1 ×5C1 ×4C1 =7×6×5×4 = 840

So there are 840 words if no letters can be repeat in four letter word.

Solution(C)

Each word begin with letter A

So no. Of 4 word letters = 1×7C1 ×7C1×7C1 = 7×7×7 = 343words

So we can made 343 words with 4 letter when Letters starts with A.

Solution(d)

Letter C must be end

7C1×7C1×7C1×1 = 343 ways

So there are 343 words which can be made for 4 letters ended with C.

Solution(e)

The second letter must be a vowel

In this we have just two vowel I.e. A,E

So no. Of ways word can be made = 7C1 ×2C1 ×7C1×7C1 = 686

So 686 words can be made with second letter must be a vowel.

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