Question

If $4$ integers are to be selected from {$1,2,3,......20$} such that the sum of the integers should be the multiple of $4$, then number of ways to select the numbers are

• A. $1220$
• B. $1120$
• C. $1200$
• D. $970$

Answer 1

The correct option is A $1220$

Every integer in the set is of the form $(4p,4p+1,4p+2,4p+3)$
We can either select 4 integers of the same type say $4p$ then no. of ways are = ${}^{\frac{20}{4}}{C}_4$
same is possible for every type
no. of ways = $4{(}^{\frac{20}{4}}{C}_4)=4{(}^5{C}_4)=20$

anothers ways is to select two integers of $4p$ type and two integers of $(4p+2)$ type then no. of ways are
= ${}^5{C}_2\times {}^5{C}_2={(}^5{C}_2{)}^2=100$
and to select two integers of $(4p+1)$ type and two integers of $(4p+3)$ type
no. of ways are = ${}^5{C}_2\times {}^5{C}_2={(}^5{C}_2{)}^2=100$

One more way is to select two integers of $4p$ type and one integer of $(4p+1)$ type and one integer of $(4p+3)$ then no. of ways are
= ${}^5{C}_2\times {}^5{C}_1\times {}^5{C}_1=250$
Also to select two integers of $(4p+2)$ type and one integer of $(4p+1)$ type and one integer of $(4p+3)$ no. of ways are
= ${}^5{C}_2\times {}^5{C}_1\times {}^5{C}_1=250$
​​​​​Also to select two integers of $(4p+3)$ type and one integer of $\left(4p\right)$ type and one integer of $(4p+2)$ no. of ways are
= ${}^5{C}_2\times {}^5{C}_1\times {}^5{C}_1=250$
Also to select two integers of $(4p+1)$ type and one integer of $(4p+2)$ type and one integer of $\left(4p\right)$ no. of ways are
= ${}^5{C}_2\times {}^5{C}_1\times {}^5{C}_1=250$
​
So total no. of ways are = $20+100+100+250+250+250+250=1220$