Question

If repetition of digits is not allowed, find how many numbers of
(i) three digits can be formed using the digits 0, 1, 2, 3, 4, 5, 6.
(ii) four digits can be formed using the digits 0, 1, 2, 3, 4. 5.
step by step explanation please​

Answer 1

Answer:

Class 11th,

Permutations and combinations

Step-by-step explanation:

First question

• There are only three digits allowed.
• We have 7 numbers.
• Repetition is not allowed.
• (If we assume the number in three boxes, each having only one digit), the first box can be filled in 7 ways.
• Next box can be filled in 6 ways, as one number is already used in the first box.
• And the last box can be filled in 5 boxes similarly.

Hence there are, in total,

7 X 6 X 5 = 210 numbers in total.

Second question

Similarly,

There are 6 X 5 X 4 X 3 = 360 numbers in total.

Another method

By the fundamental theorem of permutations,

• If Repetition is not allowed,
• If we have 'n' things,
• Out of which only 'r' can be used at a time,

(obviously n is greater than or equal to r)

Then the total number of permutations (arrangements)

P (n, r) = n! / (n-r)!

So, in the first question,

No. of total permutations =

$\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} \\ = 7 \times 6 \times 5 \\ = 210$

And for the second question,

No. of total permutations =

$\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} \\ = 6 \times 5 \times 4 \times 3 \\ = 360$

Notice

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