Question

How many number of 5 digits can be formed without repetition if: (i) 2,3 and 5 always occur in each number. (ii) 0 never occurs.

Answer 1

5 digit numbers are to be formed without repetation.

There are 5 places to be filled.

(i) if 2,3 and 5 always occur, 3 places are occupied by these numbers.

It can be arranged in 5P3 ways.

Rest two places(out of 5) can be filled by the rest 7 digits(0,1,4,6,7,8,9)

It can be done in 7P2 ways.

$Total\ ways ={}^5P_3 \times {}^7P_2 \\ \\ = \frac{5!}{(5-3)!} \times \frac{7!}{(7-2)!} \\ \\ = \frac{5!}{2!} \times \frac{7!}{5!} \\ \\ = 60 \times 42 = 2520$

But this includes the cases where 0 is in the beginning. if 0 remains in the beginning, it will be a 4 digit number. so you need to subtract it.

If we keep 0 at first place and form the numbers, total numbers will be = 1× 4P3×6 = 24×6 = 144

$So \: total \: numbers = 2520 - 144 = \boxed{2376}$

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(ii)
if zero never occurs, the five places can be filled by 9 digits (1 to 9)

Number of ways to do it = 9P5

$Total\ ways = {}^9P_5 = \frac{9!}{(9-5)!} \\ \\ = \frac{9!}{4!} \\ \\ = 9 \times 8 \times 7 \times 6 \times 5 \\ \\ = \boxed{15120}$

Answer 2

Answer:

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