wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

Number of 5 digit numbers which are divisible by 5 and each number containing the digit 5, digits being all different is equal to K4!, then the value of K is


A

84

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

168

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

188

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

208

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B

168


Finding the value of k:

The number which is divisible by 5 when the digit at unit place of that number must be either 5 or 0

Given that,

Each 5 digit numbers which are divisible by 5 must contain the digit 5

Case 1: Consider 0 at unit place

Number of 5 digit numbers containing digit 5 in this case =4×8×7×6=1344

Case 2: Consider 5 at unit place

First digit cannot be the 0.

Number of 5 digit numbers in this case =8×8×7×6=2688

Therefore, total number of 5 digit numbers which are divisible by 5 must contain the digit 5=1344+2688=4032

Now,

K4!=4032K4!=168×24K4!=1684!

Equating both side

K=168

Hence, the value of K=168


flag
Suggest Corrections
thumbs-up
16
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Permutations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon