In how many ways can a six digit number(no repetition) be formed such that sum is odd.

In how many ways can a six digit number (no repetition) be formed such that sum is odd.

My approach: It can be divided into the following cases, 1 odd and 5 even, 3 odd and 3 even, 5 odd and 1 even.

Second approach: $$\frac{\binom{10}{6}\cdot6!}{2}$$ Is it correct, or is there an easier alternative?

• If you have two odd numbers and four even numbers... what is the sum? – JMoravitz Jan 21 '17 at 7:41
• Yes, apologies, will strike it out. – user371530 Jan 21 '17 at 7:42
• I'm guessing you aren't allowed a leading zero – Joffan Jan 21 '17 at 7:45
• Im aware of elementary combinatorics, how can we arrange for that condition? – user371530 Jan 21 '17 at 7:47

Since $\frac9{10}$ of the numbers will not have leading zeroes, the simplest way is to multiply your second formulation by $0.9$, thus

answer $= \dfrac{^{10}C_6\cdot6!\cdot 0.9}{2},\;\;$ or more succinctly, $\;\;0.45\times^{10}P_6$

Your second approach is including $6$ digit numbers which start with $0$.

Remove numbers having $0$ in their starting place. $$= \frac{^9C_5\cdot5!}{2}$$

Total ways: $$= \frac{^{10}C_6\cdot6!}{2}- \frac{^9C_5\cdot5!}{2}$$

The another approach for verifying would be counting the numbers which don't have $0$ in them.$$= \frac{^9C_6\cdot6!}{2}$$

And then also considering the numbers which have $0$ but not in starting place. We can place $0$ in five positions(i.e. $\times\ \_\ \_\ \_\ \_\ \_\$) and the remaining $5$ digits can be arranged in leftover five positions.$$= \frac{5\cdot\ ^9C_5\cdot5!}{2}$$

Total number of ways is $$= \frac{^9C_6\cdot6!}{2}+ \frac{5\cdot\ ^9C_5\cdot5!}{2}$$

Hint -

You have to solve it for different cases and then add them.

Case 1 - 5 are even and 1 odd.

Case 2 - 3 are even and 3 are odd.

Case 3 - 1 is even and 5 are odd.

Also second method you are using is not correct. As includes numbers with starting digit 0, case with all digits even and all digits odd.