Question

The number of six digit numbers which have 3 digits even & 3 digits odd, if each digit is to be used at most once is
(A). 64800
(B). 28600
(C). 28800
(D). 36000

Answer 1

HINT: - The formula for calculating the number of ways in which r things can be chosen from n different things is as follows
$={}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
The formula for arranging n different things is as follows
$=n!$
An important thing that is to be taken care of is that for making the number a 6 digit number then 0 or zero cannot come at the first place or the highest position as it will make the number a 5 digit number.

Complete step-by-step solution -
As mentioned in the hint, we will first find the number of cases irrespective of the position of zero as follows
Now, using the formula that is given in the hint, we get
$={}^{5}{{C}_{3}}\cdot {}^{5}{{C}_{3}}\cdot 6!$
(As we will choose 3 even numbers out of 5(0,2,4,6,8) and 3 odd numbers out of 5(1,3,5,7,9) and then we will arrange them regardless of the position of zero)
Now, if we fix zero at the highest position, we can write the total numbers that will hence form are as follows
$={}^{5}{{C}_{3}}\cdot {}^{4}{{C}_{2}}\cdot 5!$
(As we will choose 2 even numbers out of 4 (2,4,6,8) and 3 odd numbers out of 5 (1,3,5,7,9) and then we will arrange them except moving zero from its position)
Now, we will get the required number of numbers as follows

& =\left( {}^{5}{{C}_{3}}\cdot {}^{5}{{C}_{3}}\cdot 6!-{}^{5}{{C}_{3}}\cdot {}^{4}{{C}_{2}}\cdot 5! \right) \\\ & =\left( 100\cdot 6!-60\cdot 5! \right) \\\ & =10\cdot 5!\left( 60-6 \right) \\\ & =540\times 120 \\\ & =64800 \\\ \end{aligned}$$ Hence, this is the number of the numbers that can be formed. NOTE: - The key concept is to select the digits according to the given conditions which are very essential to getting the right answer. Sometimes students make mistakes by using the direct formula of permutation, sometimes it is difficult to find an answer by using permutations.