Question

# The number of numbers can be formed by taking any 2 digits from digits 6,7,8,9 and 3 digits from 1, 2, 3, 4, 5 is -

^{5}C_{3 }× ^{4}C_{2} × 3! × 2!
^{5}P_{3} × ^{4}P_{2} × 5!
^{5}C_{3} × ^{4}C_{2} × 5!
^{5}C_{3} × ^{4}C_{2} ×

^{5}C_{3 }×^{4}C_{2}× 3! × 2!^{5}P_{3}×^{4}P_{2}× 5!^{5}C_{3}×^{4}C_{2}× 5!^{5}C_{3}×^{4}C_{2}×Hint:

### No repetition is allowed.

Number of ways of taking any 2 digits from digits 6,7,8,9 = similarly find out the rest.

## The correct answer is: ^{5}C_{3} × ^{4}C_{2} × 5!

### We have to form a 5digit number by taking any 2 digits from digits 6,7,8,9 and 3 digits from 1, 2, 3, 4, 5

Number of ways of taking any 2 digits from digits 6,7,8,9 =

Number of ways of taking any 3 digits from 1, 2, 3, 4, 5 =

Number of ways of forming a 5 digit number = 5!

Thus, the number of numbers can be formed by taking any 2 digits from digits 6,7,8,9 and 3 digits from 1, 2, 3, 4, 5 is - 5! .

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