Advertisements
Advertisements
Question
How many 6-digit telephone numbers can be constructed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each numbers starts with 35 and no digit appear more than once?
Advertisements
Solution
Given that each number starts with 35. We need a 6 digit number. So we have to fill only one’s place, 10’s place, 100th place, and 1000th places. We have to use 10 digits.
In these digits, 3 and 5 should not be used as a repetition of digits is not allowed. Except for these two digits, we have to use 8 digits. One’s place can be filled by any of the 8 digits in different ways, 10’s place can be filled by the remaining 7 digits in 7 different ways.
100th place can be filled by the remaining 6 different ways and 1000th place can be filled by the remaining 5 digits in 5 different ways.
∴ Number of 6 digit telephone numbers = 8 × 7 × 6 × 5 = 1680
APPEARS IN
RELATED QUESTIONS
In how many ways can 4 letters be posted in 5 letter boxes?
The number of words from the letters of the word 'BHARAT' in which B and H will never come together, is
The number of ways in which 6 men can be arranged in a row so that three particular men are consecutive, is
The product of r consecutive positive integers is divisible by
The number of words that can be made by re-arranging the letters of the word APURBA so that vowels and consonants are alternate is
The number of permutation of n different things taken r at a time, when the repetition is allowed is ______.
Determine the number of permutations of the letters of the word SIMPLE if all are taken at a time?
How many strings can be formed from the letters of the word ARTICLE, so that vowels occupy the even places?
Let b1, b2, b3, b4 be a 4-element permutation with bi ∈ {1, 2, 3, .......,100} for 1 ≤ i ≤ 4 and bi ≠ bj for i ≠ j, such that either b1, b2, b3 are consecutive integers or b2, b3, b4 are consecutive integers. Then the number of such permutations b1, b2, b3, b4 is equal to ______.
If m+nP2 = 90 and m–nP2 = 30, then (m, n) is given by ______.
