Hello. This lesson will cover a few examples to help you understand better the fundamental principles of counting. Nội dung chính
Example 1 Find the number of 3-digit numbers formed using the digits 3, 4, 8 and, 9, such that no digit is repeated.
Solution The ‘task’ of forming a 3-digit number can be divided into three subtasks – filling the hundreds place, filling the tens place and filling the units place – each of which must be performed to complete the task.
Now there are 4 ways to complete the first subtask [i.e. filling the thousands place], since we have 4 digits to begin with.
Next, there are 3 ways to complete to fill the tens place [the second subtask], as now there are only 3 digits left to be used [repetition of digits isn’t allowed]. Lastly, there are 2 ways to complete the third task.
Therefore the total number of ways to complete the task [which equals the required number of 3-digit numbers] will be 4 x 3 x 2 = 24. Pretty neat.
I’ll list out the 24 numbers, in case you don’t believe me: 348 349 384 389 394 398 438 439 483 489 493 498 834 839 843 849 893 894 934 938 943 948 983 984
Note that the digits could have been filled in any order, not necessarily in the order which I did. The answer obtained would have been the same.
Example 2 How many 3-digit numbers can be formed using the digits 4, 5 and 6, such that the digit can repeat?
Solution This seems similar. But this time the number of ways to fill the hundreds, tens and the units place would be 3 in each case, [and not 3, 2, 1] because the digits are allowed to be repeated.
Therefore the number of such 3-digit numbers would be 3 x 3 x 3 = 27. I’ll list them out again.
444 445 446 454 455 456 464 465 466 544 545 546 554 555 556 564 565 566 644 645 646 654 655 656 664 665 666
Okay. Let’s move on to a little more complex problems. I’ll stick to counting numbers in this lesson.
Example 3 Find the number of 5-digit even numbers formed using the digits 1 to 9, such that repetition of digits is
[i] allowed
[ii] not allowed
Solution I suggest that you grab a pen / paper and try to compute the answer yourself before reading further.
[i] In this case we have a restriction on filling the units place, as the number needs to be even. There are only 4 ways to do so [using 2, 4, 6 or 8].
For the other four places, we have 9 choices each [as repetition is allowed].
Therefore the number of ways will be 9 x 9 x 9 x 9 x 4 = 32805 [I won’t list the numbers this time. You have no option but to believe me.]
[ii] This one is a little tricky. You might say that the answer is 9 x 8 x 7 x 6 x 4, as there will be 4 ways to fill the last digit [using 2, 4, 6 or 8].
But there’s a problem. We’re not sure that after filling the first four places [from the left], whether we’ll be left with 4 choices for the units place or not. Because one of the even digits might have already been used before we reach the units place.
To get around this problem, we fill the units place first using the even digits, followed by 8, 7, 6 and 5 choices respectively for the remaining places.
Therefore the number of such 5-digits numbers will be 4 x 8 x 7 x 6 x 5 = 6720.
Example 4 Find the number of 4-digit odd numbers formed using the digits 0 to 9 such that repetition of digits is
[i] allowed
[ii] not allowed
Solution The units digit in this case can be filled in 5 ways [using the odd digits]. Now, there are only 9 ways to fill the thousands place, as 0 cannot be used there. The remaining places can be filled 10 ways each.
Therefore the total numbers in this case would be 9 x 10 x 10 x 5 = 4500.
[Note that this is half of the total 9000 4-digit numbers possible, the other half being even]
[ii] We’ll first fill the units place with an odd digit – 5 ways. Next we have the thousands digit – 8 ways [leaving out the 0].
Finally, 8 ways for the hundreds place [the 0 is back] and 7 ways for the tens place. The total number will be 5 x 8 x 8 x 7 = 2240
I guess that’ll be it for this lesson. As a practice problem try and count the number of 5-digit even numbers using the digits 0 to 9, with and without repetition.
The next lesson will establish a few formulae / expressions to deal with specific situations. See you there!
Answer : 240
Solution : We have 6 digits, viz., 5,6,7,8,9, and 0 and we have to form numbers
[integers] greater than 6,00,000, which are odd.
So the first place [lakh's position] should be `ge 6` and the last position [i.e., unit] must be odd, i.e., 5,7, or 9.
[a] When repetitions is allowed.
First place can be filled by 6,7,8, or 9 i.e., in 4 ways, last place can be filled by 5,7, or 9 i.e., in 3 ways and each of the remaining 4 places [i.e., 2nd, 3rd, 4th, 5th] can be filled by any of the 6 digits in 6 ways. Hence, the total numbers will be
`4xx6xx6xx6xx6xx3=15552`.
The total number of numbers that can be formed is
`4xx6xx6xx6xx6xx3=15552`.
[b] When repetitions is not allowed.
Since we have restrictions on first and last places and no digit can be repeated, we have the following algorithm: 1st place having numbers `ge 6` and last place having odd numbers.
Thus, the first and the last place can be filled in 10 ways, and the remaining 4 places can be filled by the remaining 4 digits in 4!=24 ways. Hence, the total number of numbers that can be formed is `10xx24=240`.
How many 3 digit numbers can be formed using the digits 5 7 8 and 9 if repetition of digits is not allowed?
The number 215 is different from 125. There are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 if no repetition is allowed.
How many 3 digits number can be formed using digits 1 5 7 and 9?
Solution : 357, 375, 537, 573, 735, 753. Therefore, '6' three-digit numbers can be formed.
Hence, the correct option is [d]. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
How many 3 digit odd numbers can be formed from the same group of numbers of digits Cannot be used more than once?
But the ten's and hundred's place can be filled up by any of the 6 given digits in 6 ways
each.
Hence, the number of 3-digit odd numbers that can be formed =`3xx6xx6=108. `
How many three digit odd numbers can be formed by using the digits 5 6 7 8 of the repetition of digits is allowed?
Hence, by the fundamental principle of multiplication, the required number of odd numbers `= [3xx6xx6] = 108. `