The number of integers, greater than 7000 that can be formed, using the digits 3, 5, 6, 7, 8 without repetition, is [2023]
168
120
220
48
(1)
5 digits numbers = 5! = 120
Four digit number greater than 7000 = 2 × 4 × 3 × 2 = 48
Total number greater than 7000 = 120 + 48 = 168
The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is [2023]
72
120
6
12
(1)
Number of numbers lies between 5000 and 10000 are 4 digit numbers greater than 5000 so, the first digit can be filled in 3 ways (using 5, 7, 9) and as repetition is not allowed, the other choices are 4, 3, 2.
Total number of number = 3 × 4 × 3 × 2 = 72
The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is [2023]
86
89
79
84
(2)
The letters in sequence of the word ‘OUGHT’ in alphabetical order as follows: G, H, O, T, U
Word starts with G can arrange in 4! ways.
H can arrange in 4! ways.
O can arrange in 4! ways.
TG can arrange in 3! ways.
TH can arrange in 3! ways.
TOG can arrange in 2! ways.
TOH can arrange in 2! ways.
TOUGH can arrange in 1! way.
So we must have serial number of the word is 89.
The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48, is [2023]
400
507
432
472
(3)
We know that we have total 3-digit numbers only 900.
Out of these 900, the number divisible by 3
In the same way, the numbers divisible by 4
So, divisible by 3 and 4 = 75
Number divisible by either 3 or 4
Now, we have to remove the numbers divisible by 48, 144, 192 ....... 18 terms.
Hence, required numbers, who are divisible by either 3 or 4 but not by 48 is
The number of ways of selecting two numbers and and such that 2 is the remainder when is divided by 23 is [2023]
54
108
268
186
(2)
The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ________ . [2023]
(4898)
Given digits are 1, 2, 3, 4, 5, 6, 7
Total number of permutations = 7!
Let A = number of numbers containing string 153
Let B = number of numbers containing string 2467
Required number of permutations = Total number of permutations -
The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to _________ . [2023]
(26664)
Sum of all unit place numbers
In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat, is __________ . [2023]
(44)
Dearrangement of 5 students
The number of seven digit positive integers formed using the digits 1, 2, 3 and 4 only and sum of the digits equal to 12 is __________ . [2023]
(413)
Case 1:
Number of ways = [Since, 1 repeats five times]
Case 2:
Number of ways = [Since, 1 repeats 4 times and 3 repeats 2 times]
Case 3:
Number of ways = [Since, 2 repeats 3 times and 1 repeats 3 times]
Case 4:
Number of ways = [Since, 2 repeats 5 times and 1 repeats 2 times]
Case 5:
Number of ways = [Since, 1 repeats 4 times and 2 repeats 2 times]
Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits 1, 2, 3, 4, 5 with repetition, is ________ . [2023]
(16)
Total 3-digit numbers formed by 1, 2, 3, 4, 5 when the last digit in 4 and divisible by 6.
[IMAGE 18]
When last digit is 2:
[IMAGE 19]