I really forgot that we could solve it using combination and permutation.]]>

(i) {4}

(ii) {3,1}

(iii) {2,2}

(iv) {2,1,1}

(v) {1,1,1,1}

We take each case in turn.

(i) The only possible 8-digit number is 40000000.

(ii) The leading digit must be 1 or 3, and the other digit can be placed in any of the other 7 places. Thus there are 7 + 7 = 14 such 8-digit numbers.

(iii) One 2 is the leading digit and the other 2 can be placed in any of the other 7 places, so number of such 8-digit numbers is 7.

(iv) If the leading digit is 2, the two 1s can be placed in the other places in [sup]7[/sup]C[sub]2[/sub] = 21 ways. If the leading digit is 1, the other two digits can be placed in the other places in [sup]7[/sup]P[sub]2[/sub] = 42 ways. ∴ Number of such 8-digit numbers = 21 + 42 = 63.

(v) One of the 1s is the leading digit and the other 3 can be placed in the other places in [sup]7[/sup]C[sub]3[/sub] = 35 ways.

Hence the total number of such 8-digit numbers is 1 + 14 + 7 + 63 + 35 = 120.

]]>You are correct, 120 is the answer.

The answer is done using generating functions but can also be done playing spot the pattern.

]]>I am confused as I got the result as 120 and some of my friends told me they got 149.

Is their any formula to find it.

pls help.

Niharika

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