Q. 275.0( 1 Vote )

# How many different numbers of six digits can be formed from the digits 3, 1, 7, 0, 9, 5 when the repetition of digits is not allowed?

In the question, we have to find the possible number of 6 digit numbers formed by the numbers 0, 1, 3, 5, 7, 9 when repetition of digits is not allowed.

We will use the concept of multiplication because there are six sub jobs dependent on each other because a number appearing on any one place will not appear in any other place.

Since zero cannot be used in the first position from the left because then it will become a 5 digit number , so we have total of 6 numbers to choose from, but we exclude zero , so we have 5 numbers to choose for the first place , and out of 5 only one number gets occupied , so 4 numbers are left but we had only 5 choices out of 6 and zero can come on the second position from the left side , so remaining choices for the second position are 5 , the number of choices will decrease by one as we keep on going right side.

The number of ways in which we can form six digit numbers with the help of given numbers is 5 × 5 × 4 × 3 × 2 × 1 = 5 × 5! = 120 × 5 = 600

The numbers occurring on first place from left have 5 choices because zero cannot come here and when one number is placed then number occurring on second place from left should have 4 choices but now zero is included in our choice because of the second position , so there will be 5 choice instead of 4 , and so on one fewer choice will be available to every next place until one occurs.

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