8. How many words can be formed with letters H, E, A, R, T where ”E” and ”A occupy the end places?
9. There are 6 people who will sit in a row but out of them Rachel will always be at right of Sol? How many arrangements can be done?
10. License plates are formed using three letters followed by a four- digit number without repetition of either letters or digits. Zero may be chosen as the first digit of the number. How many license plates can be formed under this pattern?
Given : letters H, E, A, R, T
To Find : How many words can be formed with letters H, E, A, R, T where ”E” and ”A occupy the end places
Solution:
Assuming no repetition allowed and all letters to be used
H , R and T can be arranged in 3! = 6 ways
E and A can be arranged in 2 ways
Hence 6 * 2 = 12 words can be formed
6 peopled
2 people 1 group
so total 4 + 1 = 5
Hence 5! = 120 ways
license plates can be formed = ²⁶P₃*¹⁰P₄
= 26 * 25 * 24 * 10 * 9 * 8 * 7
= 7,86,24,000
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8) Assuming no repetition allowed and all letters to be used
H, R and T can be arranged in 3! = 6 ways
E and A can be arranged in 2 ways
Hence 6*2 = 12
6 peopled
2 people 1 group
so total 4 + 1 = 5
Hence 5! = 120 ways
license plates can be formed = 26P3*10P4
= 26 * 25 * 24 * 10* 9*8*7
= 7,86,24,000.
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9) There are 6!=720 ways to arrange 6 people in a row.
Assuming they don’t necessarily have to sit right next to each other, in half of these 720 Sol is on the right of Rachel and in the other half Rachel is on the right of Sol. So 720/2 = 360 is your answer.
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10) We will split this into two parts: the letters part and the digits part.
Letters: For the first letter, we have 26 options (assuming we’re talking about the A to Z system and not Greek letters or anything like that). For the second, we have 25 options, because we can’t use the one letter we just used. By the same logic, we have 24 ways to choose the third letter. The number of ways to choose our three letters is 26x25x24 = 15600.
Digits: For the first digit we have 10 options: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. For the second digit, we have 9 choices (we can’t repeat, so we can’t use the digit we just used), for the third digit we have 8 choices, and for the fourth digit, we have 7. The number of ways we can choose our sequence of four digits is 10x9x8x7 = 5040.
To get your final answer, just multiply! You have 15600 letter, and for each one you have 5040 possible number sequences. Your answer is thus 15600×5040=78624000
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