Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Write the number of ways in which 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys ?
The number of arrangements of the letters of the word BHARAT taking 3 at a time is
English alphabet has 11 symmetric letters that appear same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X and Y. How many symmetric three letters passwords can be formed using these letters?
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Evaluate the following.
`(3! xx 0! + 0!)/(2!)`
The possible outcomes when a coin is tossed five times:
Ten different letters of alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have atleast one letter repeated is ______.
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 ______.
The number of permutations by taking all letters and keeping the vowels of the word ‘COMBINE’ in the odd places is ______.
