Latest Elitmus Aptitude Question SOLUTION: Q. A n-digit number is to be formed using the digits {1,2,3....upto m} such that the number contains atleast one "1" and one "m". How many such numbers are possible? Ex - if n=4,

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(#M40031009) ELITMUS Numerical Ability question Keep an EYE

Q. A n-digit number is to be formed using the digits {1,2,3....upto m} such that the number contains atleast one "1" and one "m". How many such numbers are possible?

Ex - if n=4, m=6
1236,1634,5312 are amonsgt the desired numbers.repition is allowed

Asked In Elitmus ravi (11 years ago)

Unsolved Read Solution (11)

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Ajay Sharma  (11 years ago)

Ans:- {nC2 * 2! * m^(n-2)}
Sol:- Placing one '1' and one 'm' in this 'n' digit number can be done in (nC2 * 2!) ways now after placing a '1' and a 'm' we are now left with (n-2) places on this 'n' digit number.So these (n-2) places now can be occupied in m^(n-2)ways.
So total such numbers possible are :- (nc2 * 2!)*(m^(n-2))

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Karan  (11 years ago)

Is 5312 also allowed for n=4, m=6 as given in example ?
I think it does not satisfy the condition that atlease one 'm' should be there .

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Karan  (11 years ago)

such possible numbers are 2*(m^(n-2))
if n=4, m=6
no. of such possible numbers = 2*6^(4-2)=72

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puneet khandelwal  (11 years ago)

m^(n-2)first check it for some numbers

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ravi  (11 years ago)

ya its 5316

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N Arun Singh  (11 years ago)

2*9^m-2..

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ashes  (11 years ago)

356 is the answer

remaining numbers 2,3,4,5=4 numbers.

case 1:when 1 and 6 are both 2 times
1166--->4!/2!
case 2:when 1 and 3 are both 3 times
1116--->4!/3!*2
case 3:when 1 and 3 are both 1 time and other digits appear only once
16_ _ ---->4c2*4!
case 4:when 1 and 3 are both 1 time and one digit appears two time
16_ _ ----->4c1*4!/2!

adding all the cases
we get 356

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N Arun Singh  (11 years ago)

@ashes wat if m=9 and a nine digit number ??...

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kumar madhav  (11 years ago)

m-2Cn-2 (n-2)!+2

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Kushal Arora  (11 years ago)

I think it should be... [ n! * m^(n-2) ]

possibilities=> 1 m (m ways) (m ways) .... (m ways)
n digits => - - - - .... -
(nth digit)

total possibilities => 1*1*m*m*m ... upto n digits => m^(n-2)
and n-digits can be permuted as well.. so n!

total answer according to me.... n! * m^(n-2)

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T.VIGNESWARA REDDY  (10 years ago)

we form n digit number i.e 1...................n selecting two places and arranged is nP2 given 1 and m must be used so the remaining n-2 numbers can be select from m-2 numbers(1 and m is already taken) is (n-2)C(m-2) so ans is nP2*(n-2)C(m-2)*n!(and all numbers can be arranged in n! ways) nP2=n*n-1 nCr=n!/(n*(n-1))

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