Question

Mikey has 10 beads on a wire. He separates the beads into 4 groups to make a 4 digit number. The number of beads in each group gives the number of thousands, hundreds, tens and ones in the number. There must be at least one bead in each group. How many 4 digit numbers can Mikey make in this way?

Answer 1

We have to find four digits so that,

• each of these is integer,
• each of these is at least one,
• sum of these is 10.

Thus the given problem is similar to the following one.

We have to find number of integral solutions of the equation such that .

We take,

• 
• 
• 
• 

Thus our previous problem converts to find number of integral solutions of the equation such that .

We have to get 4 non-negative integers whose sum is 6.

So, number of such combination is given by

Hence, Mikey can make 84 numbers consisting four digits under the given conditions.